Best way of evaluating limits of multi variable functions

[pivoxa15]

I find that the best way of evaluating (non obvious) limits of multivariable functions is by using polar coords. First (if necessary) convert or reassign the function so that at the limit point, the variables all tend to 0 and than

sub in x=rcos(angle), y=rsin(angle) where angle is arbitary.

Than let x,y->0 as r->0. This way the limit if it exists will be found everytime because the arbitary angle will mean that graphs could come in at all possible way.

Other ways of evluating these limits involving rearranging the function and than apply the Sandwich theorm or other ones to find the limit. Or sub different functions as they near the limit point such as linear, quadratic, hyperbola etc. These ways all involve a number of steps and involve so much extra work and guess work. Polar coordinates seem to solve the problem in one go. My Uni does not teach the polar coords way but the latter ways which is strange to me.

 

[benorin]

The method used depends first on whether or not you think the limit exists: if it does not exist (dne), then the method of subing-in different functions to show that the value of the limit varies depending on how the point is approached, example:

If $f(x,y)=\frac{xy}{x^2+y^2},$ then $\lim_{(x,y)\rightarrow (0,0)} f(x,y)$ is dne since

$$\lim_{x\rightarrow 0} f(x,\pm x) = \lim_{x\rightarrow 0} \frac{x(\pm x)}{x^2+(\pm x)^2}= \lim_{x\rightarrow 0} \frac{\pm x^2}{2x^2} =\pm \frac{1}{2},$$

which implies that the value of the limit depends on the method of approach [end example.]

Also, if after transforming to polar/sphereical coordinates (when the limit is at the origin) the function contains $$\theta$$ and/or $$\phi$$, then it (the limit) is dne.

Otherwise, that is when the limit does seem to exist, clever use of inequalities in conjunction with the $$\epsilon ,\delta$$ definition of a limit may be employed to produce a proof; see this thread (read post #7) for an example of this technique.

 

[pivoxa15]

The method used depends first on whether or not you think the limit exists: if it does not exist (dne), then the method of subing-in different functions to show that the value of the limit varies depending on how the point is approached, example:

If $f(x,y)=\frac{xy}{x^2+y^2},$ then $\lim_{(x,y)\rightarrow (0,0)} f(x,y)$ is dne since

$$\lim_{x\rightarrow 0} f(x,\pm x) = \lim_{x\rightarrow 0} \frac{x(\pm x)}{x^2+(\pm x)^2}= \lim_{x\rightarrow 0} \frac{\pm x^2}{2x^2} =\pm \frac{1}{2},$$

which implies that the value of the limit depends on the method of approach [end example.]

Also, if after transforming to polar/sphereical coordinates (when the limit is at the origin) the function contains $$\theta$$ and/or $$\phi$$, then it (the limit) is dne.

Otherwise, that is when the limit does seem to exist, clever use of inequalities in conjunction with the $$\epsilon ,\delta$$ definition of a limit may be employed to produce a proof; see this thread (read post #7) for an example of this technique.

The thread you referred to is where I first learned this method of evaluating these limits. Delta and Epsilon proofs could be used but why bother when there is one system that does everything for you. Polar cords will tell you if a limit exists and what it is. It also accutrately tell you if a limit doesn't exist when you are left with angles only. It looks like the best.

I wonder if it has any limitations.

 

[VietDao29]

Also, if after transforming to polar/sphereical coordinates (when the limit is at the origin) the function contains $$\theta$$ and/or $$\phi$$, then it (the limit) is dne.

Uhmm, not quite correct, if the angle, i.e $$\theta$$ appears in the numerator, then the limit can still exists. But if it appear in the denominator, then you should beware as the limit may not (I just say may not) exist there.
-----------------
Example 1:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 2y}{x ^ 2 + y ^ 2}$$. After changing to polar-coordinate, we have:
$$\lim_{r \rightarrow 0} r \cos ^ 2 \theta \sin \theta$$. Now since:
$$-1 \leq \sin \theta , \cos \theta \leq 1$$, so by using the Squeeze Theorem, one can show that the limit is 0, i.e:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 2y}{x ^ 2 + y ^ 2} = 0$$.
The function contains $$\theta$$ but the limit does exist.
Example 2:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{xy ^ 3}{2x ^ 2 + 3y ^ 6}$$.
By changing to polar-coordinate, we have:
$$\lim_{r \rightarrow 0} \frac{r ^ 2 (\cos \theta \sin ^ 3 \theta)}{2 \cos ^ 2 \theta + 3 r ^ 4 \sin ^ 6 \theta}$$
$$\theta$$ appears in the denominator, you cannot conclude anything about this limit. And you start to suspect that the limit does not exist, since if $$\theta \rightarrow \frac{\pi}{2}$$, as $$r \rightarrow 0$$, then the limit is in one of the Indeterminate forms 0 / 0.
And so, you should look for an example that shows the limit avaluates to different values as we approach (0, 0) along different paths.
x = 0, and x = y³ works in this case.
Example 3:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{xy}{|x| + |y|}$$. Now change to polar-coordinate, we have:
$$\lim_{r \rightarrow 0} \frac{r \sin \theta \cos \theta}{|\cos \theta| + |\sin \theta|}$$.
One can show that:
$$1 \leq | \cos \theta | + | \sin \theta | \leq \sqrt{2}$$, so by using the Squeeze Theorem, one can again, say that the limit is 0.
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{xy}{|x| + |y|} = 0$$.
So it's not that if the expression contains $$\theta$$, or if $$\theta$$ is in the denominator then the limit does not exist, it's just one should make careful judgement to see if the limit exists or not exists. And you can improve your skills by practising...
By the way, as you said before, the limitation of polar coordinate is that it not taught in your university...
--------------
P.S, and also sometimes, we need to use some well-known limits of one variable function.
Example 4:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$
In this case, polar-coordinate fails. Since if we change everything to polar-coordinate, we have:
$$\lim_{r \rightarrow 0} \frac{r (\sin \theta + \cos \theta)}{\sin (r \cos \theta) + \sin (r \sin \theta)}$$, and this looks way too complicated, right? So we must find another way to go about this problem:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$
$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{2 \frac{x + y}{2}}{2 \sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}$$
$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}$$
We then can use:
$$\lim_{r \rightarrow 0} \frac{\sin r}{r} = 1$$, and conclude that:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y} = 1$$

 

[benorin]

Quite true, but this is easily modified:

If the value of the limit depends on $$\theta$$ and/or $$\phi$$, the the limit is dne.

Note that, by the definition of a limit, we have

$$\lim_{(x,y)\rightarrow (a,b)}f(x,y)=L \Leftrightarrow \lim_{(u,v)\rightarrow (0,0)}f(u+a,v+b)=L\Leftrightarrow\lim_{r\rightarrow 0^+}f(a+r\cos\theta,b+r\sin\theta)=L$$​

according to my quick'n'dirty scratch work; but if, rather, I'm full of it: let me know.

 

[pivoxa15]

Uhmm, not quite correct, if the angle, i.e $$\theta$$ appears in the numerator, then the limit can still exists. But if it appear in the denominator, then you should beware as the limit may not (I just say may not) exist there.
-----------------
Example 1:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 2y}{x ^ 2 + y ^ 2}$$. After changing to polar-coordinate, we have:
$$\lim_{r \rightarrow 0} r \cos ^ 2 \theta \sin \theta$$. Now since:
$$-1 \leq \sin \theta , \cos \theta \leq 1$$, so by using the Squeeze Theorem, one can show that the limit is 0, i.e:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 2y}{x ^ 2 + y ^ 2} = 0$$.
The function contains $$\theta$$ but the limit does exist.
Example 2:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{xy ^ 3}{2x ^ 2 + 3y ^ 6}$$.
By changing to polar-coordinate, we have:
$$\lim_{r \rightarrow 0} \frac{r ^ 2 (\cos \theta \sin ^ 3 \theta)}{2 \cos ^ 2 \theta + 3 r ^ 4 \sin ^ 6 \theta}$$
$$\theta$$ appears in the denominator, you cannot conclude anything about this limit. And you start to suspect that the limit does not exist, since if $$\theta \rightarrow \frac{\pi}{2}$$, as $$r \rightarrow 0$$, then the limit is in one of the Indeterminate forms 0 / 0.
And so, you should look for an example that shows the limit avaluates to different values as we approach (0, 0) along different paths.
x = 0, and x = y³ works in this case.
Example 3:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{xy}{|x| + |y|}$$. Now change to polar-coordinate, we have:
$$\lim_{r \rightarrow 0} \frac{r \sin \theta \cos \theta}{|\cos \theta| + |\sin \theta|}$$.
One can show that:
$$1 \leq | \cos \theta | + | \sin \theta | \leq \sqrt{2}$$, so by using the Squeeze Theorem, one can again, say that the limit is 0.
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{xy}{|x| + |y|} = 0$$.
So it's not that if the expression contains $$\theta$$, or if $$\theta$$ is in the denominator then the limit does not exist, it's just one should make careful judgement to see if the limit exists or not exists. And you can improve your skills by practising...
By the way, as you said before, the limitation of polar coordinate is that it not taught in your university...
--------------
P.S, and also sometimes, we need to use some well-known limits of one variable function.
Example 4:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$
In this case, polar-coordinate fails. Since if we change everything to polar-coordinate, we have:
$$\lim_{r \rightarrow 0} \frac{r (\sin \theta + \cos \theta)}{\sin (r \cos \theta) + \sin (r \sin \theta)}$$, and this looks way too complicated, right? So we must find another way to go about this problem:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$
$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{2 \frac{x + y}{2}}{2 \sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}$$
$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}$$
We then can use:
$$\lim_{r \rightarrow 0} \frac{\sin r}{r} = 1$$, and conclude that:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y} = 1$$

With example 2, since (by using polar coords) you showed that different limits are possible it is clearly indeterminate. Other methods are not necessary.

How does your example 4 work with the substitution inside the sin and cos?
I agree that with this one, another method must be used since polar coords will give 0/0.

 

[VietDao29]

With example 2, since (by using polar coords) you showed that different limits are possible it is clearly indeterminate. Other methods are not necessary.

Hmm, yeah, you may say that, but giving example is also a good way.

How does your example 4 work with the substitution inside the sin and cos?
I agree that with this one, another method must be used since polar coords will give 0/0.

SInce x, and y both tend to 0, then x - y must also tends to 0, right?
So:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \cos \left( \frac{x - y}{2} \right) = 1$$, no?
Now split it into a product 2 of separate limits:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$
$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}$$
$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right)} \times \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{1}{\cos \left( \frac{x - y}{2} \right)}$$
Now since (x, y) -> (0, 0), so x + y must also tend to 0, right?
Using the well-known limit:
$$\lim_{r \rightarrow 0} \frac{\sin r}{r} = 1$$, we have:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y} = 1 \times \frac{1}{1} = 1$$.
Can you get it? :)

 

[pivoxa15]

Hmm, yeah, you may say that, but giving example is also a good way.


SInce x, and y both tend to 0, then x - y must also tends to 0, right?
So:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \cos \left( \frac{x - y}{2} \right) = 1$$, no?
Now split it into a product 2 of separate limits:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$
$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}$$
$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right)} \times \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{1}{\cos \left( \frac{x - y}{2} \right)}$$
Now since (x, y) -> (0, 0), so x + y must also tend to 0, right?
Using the well-known limit:
$$\lim_{r \rightarrow 0} \frac{\sin r}{r} = 1$$, we have:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y} = 1 \times \frac{1}{1} = 1$$.
Can you get it? :)

How did you go from:

$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$

to

$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}$$

 

[pivoxa15]

Hmm, yeah, you may say that, but giving example is also a good way.


SInce x, and y both tend to 0, then x - y must also tends to 0, right?
So:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \cos \left( \frac{x - y}{2} \right) = 1$$, no?
Now split it into a product 2 of separate limits:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$
$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}$$
$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right)} \times \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{1}{\cos \left( \frac{x - y}{2} \right)}$$
Now since (x, y) -> (0, 0), so x + y must also tend to 0, right?
Using the well-known limit:
$$\lim_{r \rightarrow 0} \frac{\sin r}{r} = 1$$, we have:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y} = 1 \times \frac{1}{1} = 1$$.
Can you get it? :)

How did you go from here:

$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$


to here:

$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}$$

 

[VietDao29]

How did you go from here:

$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$ to here:

$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}$$

pivoxa15, it's one of the Sum-to-product identities, you can look it un in your textbook or view it here.
It's:
$$\sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right)$$.
Now apply that identity, and use the fact that:
$$x + y = 2 \frac{x + y}{2}$$ to arrive at that expression.
Can you get it? :)

 

[pivoxa15]

pivoxa15, it's one of the Sum-to-product identities, you can look it un in your textbook or view it here.
It's:
$$\sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right)$$.
Now apply that identity, and use the fact that:
$$x + y = 2 \frac{x + y}{2}$$ to arrive at that expression.
Can you get it? :)

I understand. By using trig identities, you are able to evaluate the limit even when polar coords does not work.

So there are situations where polar coords fail but other methods work. Which means that if the polar coords give an indeterminate form, it does not necessary mean the limit does not exist.

 

[VietDao29]

I understand. By using trig identities, you are able to evaluate the limit even when polar coords does not work.

So there are situations where polar coords fail but other methods work. Which means that if the polar coords give an indeterminate form, it does not necessary mean the limit does not exist.

Uhmm,... it does work, but it looks way complicated than the way I show you above. The method is the same, by using the Sum to Product identities, we have:
$$\lim_{r \rightarrow 0} \frac{r (\sin \theta + \cos \theta)}{\sin (r \cos \theta) + \sin (r \sin \theta)}$$
$$= \lim_{r \rightarrow 0} \frac{r (\sin \theta + \cos \theta)}{2 \sin \left( r \frac{\cos \theta + \sin \theta}{2} \right) \cos \left(r \frac{\cos \theta - \sin \theta}{2} \right)}$$
$$= \lim_{r \rightarrow 0} \frac{r \frac{\sin \theta + \cos \theta}{2}}{\sin \left( r \frac{\cos \theta + \sin \theta}{2} \right) \cos \left(r \frac{\cos \theta - \sin \theta}{2} \right)} = 1$$.
It looks much harder, right?
By the way, you should be flexible, i.e, don't just apply polar-coordinate to solve 2 variable functions (though, it can be used on most 2-variable limit), just try every way you can think off to see if you can work out the problem.
---------------------
And again, when using polar-coordinate, the limit does not exist when the limit is dependent on $$\theta$$.
In Example 1, 3, 4, the expression is independent of $$\theta$$, i.e $$\theta$$ can take whatever value, and it does not affect the limit.
In Example 2, the expression is dependent on $$\theta$$.
Can you get this? :)

 

[pivoxa15]

Uhmm,... it does work, but it looks way complicated than the way I show you above. The method is the same, by using the Sum to Product identities, we have:
$$\lim_{r \rightarrow 0} \frac{r (\sin \theta + \cos \theta)}{\sin (r \cos \theta) + \sin (r \sin \theta)}$$
$$= \lim_{r \rightarrow 0} \frac{r (\sin \theta + \cos \theta)}{2 \sin \left( r \frac{\cos \theta + \sin \theta}{2} \right) \cos \left(r \frac{\cos \theta - \sin \theta}{2} \right)}$$
$$= \lim_{r \rightarrow 0} \frac{r \frac{\sin \theta + \cos \theta}{2}}{\sin \left( r \frac{\cos \theta + \sin \theta}{2} \right) \cos \left(r \frac{\cos \theta - \sin \theta}{2} \right)} = 1$$.
It looks much harder, right?
By the way, you should be flexible, i.e, don't just apply polar-coordinate to solve 2 variable functions (though, it can be used on most 2-variable limit), just try every way you can think off to see if you can work out the problem.
---------------------
And again, when using polar-coordinate, the limit does not exist when the limit is dependent on $$\theta$$.
In Example 1, 3, 4, the expression is independent of $$\theta$$, i.e $$\theta$$ can take whatever value, and it does not affect the limit.
In Example 2, the expression is dependent on $$\theta$$.
Can you get this? :)

Have you left out a step? It still looks like 0/0 with the numerator going to 0 (as r->0) and denominator going to 0 (0*1=0).

 

[VietDao29]

Have you left out a step? It still looks like 0/0 with the numerator going to 0 (as r->0) and denominator going to 0 (0*1=0).

Yes, of course, but have you looked at that closely? That can be done exactly the same as the way I've previouisly shown you.
I left out a step to make you do some work, and think about the problem.
You should apply the limit:
$$\lim_{t \rightarrow 0} \frac{\sin t}{t} = 1$$.
Apply that limit to this part:
$$\frac{r \frac{\sin \theta + \cos \theta}{2}}{\sin \left( r \frac{\cos \theta + \sin \theta}{2} \right)}$$.
Now, can you get it? :)

 

[hooker27]

Uhmm, not quite correct, if the angle, i.e $$\theta$$ appears in the numerator, then the limit can still exists. But if it appear in the denominator, then you should beware as the limit may not (I just say may not) exist there.
-----------------
...

Example 4:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$
In this case, polar-coordinate fails. Since if we change everything to polar-coordinate, we have:
$$\lim_{r \rightarrow 0} \frac{r (\sin \theta + \cos \theta)}{\sin (r \cos \theta) + \sin (r \sin \theta)}$$, and this looks way too complicated, right? So we must find another way to go about this problem:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$
$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{2 \frac{x + y}{2}}{2 \sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}$$
$$= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}$$
We then can use:
$$\lim_{r \rightarrow 0} \frac{\sin r}{r} = 1$$, and conclude that:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y} = 1$$

Maybe I got it wrong but it looks to me that the limit
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$
clearly does not exist, is that your result?
When approaching the (0,0) point from y=-x direction, the function is 0/0 and thus undefined (not indeterminate, as some of you say) - which means that there exists no neighbourhood of (0,0) such that
1) the function is defined on the neighbourhood.
2) the function values are "close" (closer than the gived epsilon from the limit definition) to the limit value, if any.

which is the definition of the limit. Conclusion: the limit does not exist.
Where am I wrong?

 

[pivoxa15]

VietDao, I have just realized that the function is a multivariable one either F(x,y) or F(r,theta). The limit as r->0, sin(r)/(r) -> 1 works for one variable functions because you differentiate the top and bottom to get cos(r)/1 -> 1 as r->0. But the derivative of multivariable functions are matrices and you cannot divide matrices by matrices.

So Hooker might be right in that this functions is indeterminate after all.

 

[hooker27]

VietDao, I have just realized that the function is a multivariable one either F(x,y) or F(r,theta). The limit as r->0, sin(r)/(r) -> 1 works for one variable functions because you differentiate the top and bottom to get cos(r)/1 -> 1 as r->0. But the derivative of multivariable functions are matrices and you cannot divide matrices by matrices.

So Hooker might be right in that this functions is indeterminate after all.

I believe that the limit indeed does not exist, one must be careful with multivariable limits, even with the most obvious limits. For example
$$\lim_{x \rightarrow 0} \frac{x}{x} = 1$$
but
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x}{x}$$
does not exist, for the same reasons I described in my previous post (this time when approaching (0,0) from x=0 direction - again, no neighbourhood exists...).

But, pivoxa15, when some expression is 0/0 or whatever/0 or tan(pi/2), it's value is undefined and certainly not indeterminate. There is no nonsense like indeterminate, although it is in textbooks for centuries. I am not trying to be smarter that all of you, but it's time to realize that something either is defined or it is not and there's nothing in between. Maybe I act stupid but there are some things in math against which I fight whenever I encounter them and this is one of them, so forgive me if I am overreacting.

 

[VietDao29]

Yeah, sorry for such stupid confusion... :blush:
hooker27, I don't understand:

But, pivoxa15, when some expression is 0/0 or whatever/0 or tan(pi/2), it's value is undefined and certainly not indeterminate. There is no nonsense like indeterminate, although it is in textbooks for centuries. I am not trying to be smarter that all of you, but it's time to realize that something either is defined or it is not and there's nothing in between. Maybe I act stupid but there are some things in math against which I fight whenever I encounter them and this is one of them, so forgive me if I am overreacting.

Yes, it's true that you are overreacting. whatever / 0 is not one of the Indeterminate Forms.
Let's consider:
0 / 0 = a. It's means that 0 = 0a. Which is a true statement, right? So when evaluating a limit, and you encounter 0 / 0 form, you don't instantly know what the answer is. It can be 1, 2, 0, or even infinity. Hence, it's the Indeterminate Forms.
It's not like some limits as: 1 / 2, or 3 / 4, or..., all of which you can just "plug" the value of x into get the answer. Indeterminate Forms requires you to do some manipulations before you can plug x in.

 

[hooker27]

Yes, it's true that you are overreacting. whatever / 0 is not one of the Indeterminate Forms.
Let's consider:
0 / 0 = a. It's means that 0 = 0a. Which is a true statement, right? So when evaluating a limit, and you encounter 0 / 0 form, you don't instantly know what the answer is. It can be 1, 2, 0, or even infinity. Hence, it's the Indeterminate Forms.
It's not like some limits as: 1 / 2, or 3 / 4, or..., all of which you can just "plug" the value of x into get the answer. Indeterminate Forms requires you to do some manipulations before you can plug x in.

Ok, I'll calm it down but tell me waht you think about this:

You are right that 0=0a is true for all a but 0/0=a is nonsense since 0/0 is not assigned any value, you can't divide by 0. But I don't want to argue over something so simple, let's look at this:

If you say, that 1/0 does not have any real value or for short you say that 1/0 is undefined, you just developed a name for some category of expressions (undefined = does not have any value). And then you trip over 0/0 and say "This I will call indeterminate!", I get the feeling that you imply that 0/0 is in some way different from 1/0 (in meaning, not subtraction), otherwise you would call it undefined as well. The only possible difference might be that you believe that under some special circumstances, 0/0 might have some real value. But this is not true!. Since the creation of the world 0/0 never had any value and it will never have any in the future, under whatever circumstances and regardless of how advanced mathematical weapon you bring to it (limits, for example). This is the reason why I don't like calling expressions like 0/0, 1^infinity etc. indeterminate because it means that you admit any difference between 1/0 and 0/0 etc. where there is none.

Now I understand that the reason why you tend to call such expressions indeterminate is the fact that, for example
$$\lim_{x \rightarrow 0} \frac{x}{x} = 1$$ and $$\lim_{x \rightarrow 0} \frac{2x}{x} = 2$$
and both expressions inside the limits are "0/0" so one gets the impression that the value of 0/0 chages depending on the context. But in the background of this "impression" is a big misunderstanding of limits. A limit $$\lim_{x \rightarrow a} f(x)$$ never looks and the funtion at the point a, whether or not the function has a value in a has absolutely no influence whatsover on the limit value so in the example above, the point 0/0 or 2*0/0 was never considered (evaluated), so you can't say that once it's value was 1 and then 2. And - logically going on - you can't give it a name ("indeterminate") depending on the result of the limit.

You should now understand why I am against giving weird names to expressions like 0/0, tell me what your opinion is. Or forget it forever, I don't care, it's just that I'd like to see if this is making any sense to anyone except me.

Speaking of limits, my english is pretty limited so forgive me if it's too painful to read.

 

[pivoxa15]

Ok, I'll calm it down but tell me waht you think about this:

You are right that 0=0a is true for all a but 0/0=a is nonsense since 0/0 is not assigned any value, you can't divide by 0. But I don't want to argue over something so simple, let's look at this:

If you say, that 1/0 does not have any real value or for short you say that 1/0 is undefined, you just developed a name for some category of expressions (undefined = does not have any value). And then you trip over 0/0 and say "This I will call indeterminate!", I get the feeling that you imply that 0/0 is in some way different from 1/0 (in meaning, not subtraction), otherwise you would call it undefined as well. The only possible difference might be that you believe that under some special circumstances, 0/0 might have some real value. But this is not true!. Since the creation of the world 0/0 never had any value and it will never have any in the future, under whatever circumstances and regardless of how advanced mathematical weapon you bring to it (limits, for example). This is the reason why I don't like calling expressions like 0/0, 1^infinity etc. indeterminate because it means that you admit any difference between 1/0 and 0/0 etc. where there is none.

Now I understand that the reason why you tend to call such expressions indeterminate is the fact that, for example
$$\lim_{x \rightarrow 0} \frac{x}{x} = 1$$ and $$\lim_{x \rightarrow 0} \frac{2x}{x} = 2$$
and both expressions inside the limits are "0/0" so one gets the impression that the value of 0/0 chages depending on the context. But in the background of this "impression" is a big misunderstanding of limits. A limit $$\lim_{x \rightarrow a} f(x)$$ never looks and the funtion at the point a, whether or not the function has a value in a has absolutely no influence whatsover on the limit value so in the example above, the point 0/0 or 2*0/0 was never considered (evaluated), so you can't say that once it's value was 1 and then 2. And - logically going on - you can't give it a name ("indeterminate") depending on the result of the limit.

You should now understand why I am against giving weird names to expressions like 0/0, tell me what your opinion is. Or forget it forever, I don't care, it's just that I'd like to see if this is making any sense to anyone except me.

Speaking of limits, my english is pretty limited so forgive me if it's too painful to read.

Wordings aside, I think you have to be careful with things like

$$\lim_{x \rightarrow 0} \frac{x}{x} = 1$$ and $$\lim_{x \rightarrow 0} \frac{2x}{x} = 2$$

If you were in some maths or physics problem and you come to x/x would you cancel it and write 1 and in addition put a note beside it saying x cannot equal 0?

Or would you do the cancellation to get 1 and leave it at that?

I haven't seen anyone do the former.

 

[pivoxa15]

VietDao, what do you think about this limit and my critisism the limit sin(t)/t not existing because t is a multivariable function in our case where as the limit of it being 1 only appies when t is a single variable.

$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$

 

[VietDao29]

VietDao, what do you think about this limit and my critisism the limit sin(t)/t not existing because t is a multivariable function in our case where as the limit of it being 1 only appies when t is a single variable.

$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}$$

Hmmm, I don't really understand what you mean...
But $$\lim_{r \rightarrow 0} \frac{\sin r}{r}$$ is indeed 1. You can look at your calc textbook for a proof (the one that does not use L'Hopital's rule).
-------------------
There's a flaw in my example 4, since if you approach (0, 0) along the path y = -x, the function:
$$\frac{x + y}{\sin x + \sin y}$$ is not defined at all. And hence there's no limit at (0, 0).
But consider the limit:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{|x| + |y|}{\sin (|x| + |y|)} = 1$$ (the function is defined for every x, and y in the neighbourhood of (0, 0)). Can you get this?
Sorry again for causing such confusion.

 

[VietDao29]

A limit $$\lim_{x \rightarrow a} f(x)$$ never looks and the funtion at the point a, whether or not the function has a value in a has absolutely no influence whatsover on the limit value so in the example above, the point 0/0 or 2*0/0 was never considered (evaluated), so you can't say that once it's value was 1 and then 2. And - logically going on - you can't give it a name ("indeterminate") depending on the result of the limit.

Please note that the numerator and the denominator only tend to 0, they are not 0. And you should also note that the function x / x is itself discontinuous at x = 0. But it has the limit 1 as x tends to 0.

You should now understand why I am against giving weird names to expressions like 0/0, tell me what your opinion is.

Hmm, as a matter of fact, I don't think that's a weird name, I think it's logical... :)

 

[pivoxa15]

Hmmm, I don't really understand what you mean...
But $$\lim_{r \rightarrow 0} \frac{\sin r}{r}$$ is indeed 1. You can look at your calc textbook for a proof (the one that does not use L'Hopital's rule).
-------------------
There's a flaw in my example 4, since if you approach (0, 0) along the path y = -x, the function:
$$\frac{x + y}{\sin x + \sin y}$$ is not defined at all. And hence there's no limit at (0, 0).

Didn't you say that the function does not have to exist in order for the limit to exist in your most recent post? So a function could have a limit at a place where it may not be defined.

If you do direct substitution, y=-x, it seems that the limit is 0/0 but according to you it should be 1 because of the limit as x->0 sin(x)/x=1

A proof without L'Hopital's rule would be to use the sandwich rule
abs(sinx) <= abs(x) so since lim x/x = 1 sin(x)/x must also be convergent. If x is 0 than sinx=x so given lim x/x =1, we know that sin(x)/x =1.

hence even though when you approach (0,0) along y=-x, the function is the form 0/0 it simply means that direct substitution is ineffective. The limit exists and is 1.

 

[VietDao29]

Didn't you say that the function does not have to exist in order for the limit to exist in your most recent post? So a function could have a limit at a place where it may not be defined.

Yes, the function does not have to be defined at (0, 0), but the function must be defined for some neighbourhood of (0, 0).
Let's review the definition for 2-variable function:
If
$$\forall\epsilon >0, \exists \delta >0 \mbox{ such that } 0<\sqrt{(x - \alpha)^2 + (y - \beta)^2}<\delta\Rightarrow \left| f(x, y) - L \right| < \epsilon$$
Then we say that:
$$\lim_{\substack{x \rightarrow \alpha \\ y \rightarrow \beta}} f(x, y) = L$$.
Okay, so from the definition above, the limit can still exist even if the function is not defined at $$( \alpha, \ \beta )$$.
But it must be defined for every x, and y arround $$( \alpha, \ \beta )$$, right?
So if you change the example 4 a bit, it will give 1 as the limit:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{|x| + |y|}{|\sin x| + |\sin y|} = 1$$. Prove the same, but this time, it's defined for every x, y arround (0, 0).
------------------
Another example of a limit that does not exist is:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} xy \ln |xy|$$, since if you approach (0, 0) along the x or y axis, the function is undefined, hence, it's not defines for all x, y arround (0, 0).
Can you get this? :)

 

[pivoxa15]

Yes, the function does not have to be defined at (0, 0), but the function must be defined for some neighbourhood of (0, 0).
Let's review the definition for 2-variable function:
If
$$\forall\epsilon >0, \exists \delta >0 \mbox{ such that } 0<\sqrt{(x - \alpha)^2 + (y - \beta)^2}<\delta\Rightarrow \left| f(x, y) - L \right| < \epsilon$$
Then we say that:
$$\lim_{\substack{x \rightarrow \alpha \\ y \rightarrow \beta}} f(x, y) = L$$.
Okay, so from the definition above, the limit can still exist even if the function is not defined at $$( \alpha, \ \beta )$$.
But it must be defined for every x, and y arround $$( \alpha, \ \beta )$$, right?
So if you change the example 4 a bit, it will give 1 as the limit:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{|x| + |y|}{|\sin x| + |\sin y|} = 1$$. Prove the same, but this time, it's defined for every x, y arround (0, 0).
------------------
Another example of a limit that does not exist is:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} xy \ln |xy|$$, since if you approach (0, 0) along the x or y axis, the function is undefined, hence, it's not defines for all x, y arround (0, 0).
Can you get this? :)

I agree that a neighbourhood around the limit point was be defined by the
function. Otherwise it would be impossible for the function to approach
the original limit. For example the function cannot 'jump' across an
undefined gap.

With your second example, I confirmed that the limit does not exist by
using the polar coords and got 0ln(0) which is undefined hence the limit
does not exist. It further shows the unbreakableness of using polar
coords. I wonder if there is a case where
1. Using polar coords evualtes a limit which is incorrect in that a
different value should be the correct one
or
2. Show that a limit does not exist where in fact there does exist a limit.

Back to your example, by showing that going along the single x and y
ordinates produce no limit, have you shown that going along any
combination of x and y produce no limit? The reason is that to show no
limit exists you must prove that any way of going into the origin
gives no limit, not just the single x and y ordinates. This is another
reason why using polar coords is so good. You could define theta to be any real number hence effectively you allow x and y to go into the origin in
all possible ways at once.

 

[VietDao29]

With your second example, I confirmed that the limit does not exist by
using the polar coords and got 0ln(0) which is undefined hence the limit
does not exist.

What do you mean? Note that the function can be undefined at the point you are evaluationg the limit.
Does this limit exist according to you?
$$\lim_{(x, \ y) \rightarrow (0, \ 0)} (|x| + |y|) \ln (|x| + |y|)$$
The function is defined for every x, and y arround 0. And its limit can be evaluated by:
$$\lim_{(x, \ y) \rightarrow (0, \ 0)} (|x| + |y|) \ln (|x| + |y|) = \lim_{a \rightarrow 0 ^ +} a \ln(a) = \lim_{a \rightarrow 0 ^ +} \frac{\ln a}{\frac{1}{a}} = 0$$ (Using L'Hopital's rule).

It further shows the unbreakableness of using polar
coords. I wonder if there is a case where
1. Using polar coords evualtes a limit which is incorrect in that a
different value should be the correct one
or

I can assure you this won't ever happen. :)

2. Show that a limit does not exist where in fact there does exist a limit.

This won't happen, either. :)

Back to your example, by showing that going along the single x and y
ordinates produce no limit, have you shown that going along any
combination of x and y produce no limit? The reason is that to show no
limit exists you must prove that any way of going into the origin
gives no limit, not just the single x and y ordinates. This is another
reason why using polar coords is so good. You could define theta to be any real number hence effectively you allow x and y to go into the origin in
all possible ways at once.

No, if we approach (0, 0) along the x, and y axes, the function is NOT defined. Hence, there exist no limit. Look at the definition for limit of 2 variable function in your book again. For the limit to exist, there must first be some neighbourhood arround (0, 0) that the function is defined for every x, and y in that neighbourhood.

 

[pivoxa15]

What do you mean? Note that the function can be undefined at the point you are evaluationg the limit.

I made a mistake. I got to 0log(0) and did not evaluate it further. This limit is not undefined but 0 according to what you showed in your example below. It is possible to not use L'Hopital's rule and use the sandwich rule instead which is must simpler in this case. Note (a)log(a) < a^2 and since as a->0 a^2-> 0, (a)log(a) also will go to 0 and will not be undefined as I mistakely said in my previous post.

Obviously the point (0,0) is undefined but the limit exists which is 0.

Does this limit exist according to you?
$$\lim_{(x, \ y) \rightarrow (0, \ 0)} (|x| + |y|) \ln (|x| + |y|)$$
The function is defined for every x, and y arround 0. And its limit can be evaluated by:
$$\lim_{(x, \ y) \rightarrow (0, \ 0)} (|x| + |y|) \ln (|x| + |y|) = \lim_{a \rightarrow 0 ^ +} a \ln(a) = \lim_{a \rightarrow 0 ^ +} \frac{\ln a}{\frac{1}{a}} = 0$$ (Using L'Hopital's rule).

This limit is evluated exactly like the previous example and exists which is 0.

No, if we approach (0, 0) along the x, and y axes, the function is NOT defined. Hence, there exist no limit. Look at the definition for limit of 2 variable function in your book again. For the limit to exist, there must first be some neighbourhood arround (0, 0) that the function is defined for every x, and y in that neighbourhood.

I am aware that a neighbourhood around (0,0) must be defined but even if the limit does not exist when approached from the individual x and y axes, it could exist when approached in a fancier way like a parabolic approach. There is an infinite number of ways the graph could approach the origin. That is why I think polar coords are so good but it takes account of this infinity in one go with the theta.

This is the reason why polar coords is so good because when evaluating limits, you don't have to find all sorts of different ways of approaching the origin. i.e. you don't have to think about y=kx y=kx^2 etc. You can do all of it in one hit.

 

[VietDao29]

I am aware that a neighbourhood around (0,0) must be defined but even if the limit does not exist when approached from the individual x and y axes, it could exist when approached in a fancier way like a parabolic approach.

Uhmm, this is not quite correct... :)
To prove that the limit does not exist, you can do one of the following:
1. The limit tends to +, or - infinity along some path.
2. The limit evaluates to some different values depends on which path we approach it.
3. There is no neighbourhood of (0, 0) that the function is defined for every x, and y in that neighbourhood.
------------------
The limit:
$$\lim_{(x, \ y) \rightarrow (0, \ 0)} |xy| \ln|xy|$$
does not exist. Since the function is not defined along the x, or y-axis (number 3).
The limit still does not exist even if it does evaluate to 0 as you approach it along the x = ky path, or y = kx (k is not 0), or even in a parabolic path like y = x² (Or whatever path you like )...
However, the limit $$\lim_{(x, \ y) \rightarrow (0, \ 0)} (|x| + |y|) \ln (|x| + |y|) = 0$$ (explained in the previous post). Notice that the function is defined for every pair of (x, y), but (0, 0). Right?

This is the reason why polar coords is so good because when evaluating limits, you don't have to find all sorts of different ways of approaching the origin. i.e. you don't have to think about y=kx y=kx^2 etc. You can do all of it in one hit.

Yes, this is is correct. :)
----------
Is everything clear now? :)

 

[pivoxa15]

Uhmm, this is not quite correct... :)
To prove that the limit does not exist, you can do one of the following:
1. The limit tends to +, or - infinity along some path.
2. The limit evaluates to some different values depends on which path we approach it.
3. There is no neighbourhood of (0, 0) that the function is defined for every x, and y in that neighbourhood.
------------------
The limit:
$$\lim_{(x, \ y) \rightarrow (0, \ 0)} |xy| \ln|xy|$$
does not exist. Since the function is not defined along the x, or y-axis (number 3).
The limit still does not exist even if it does evaluate to 0 as you approach it along the x = ky path, or y = kx (k is not 0), or even in a parabolic path like y = x² (Or whatever path you like )...
However, the limit $$\lim_{(x, \ y) \rightarrow (0, \ 0)} (|x| + |y|) \ln (|x| + |y|) = 0$$ (explained in the previous post). Notice that the function is defined for every pair of (x, y), but (0, 0). Right?

Yes, this is is correct. :)
----------
Is everything clear now? :)

One way to prove that a limit does not exist is to show that no matter which path one approach the point under study, no limit will ever exist.
Hence if you find one way to approach the point resulting in a defined value of the function than the limit exists at that point. Even if all other ways of approaching this point results in an undefined value.

You said that for a limit to exist than a neighbourhood around this point must exist. Do you mean a complete neighbourhood (i.e. with every point defined in it?) If so than that is not necessary because the definition is that:
'A point z is a limit point of a set S (real or complex) if any neighbourhood of z contains a point of S, other than z itself.'
Hence only one point (meaning one path into the origin) is enough to prove that a limit exists.
So with your example, even if the function is undefined going along the x or y axis, it does not mean that another path cannot approach the origin legitamatly. Therefore there might be a point around the neighbourhood of the origin that is an element of the function $$\lim_{(x, \ y) \rightarrow (0, \ 0)} |xy| \ln|xy|$$
Using polar coords or even without it, and applying the sandwich theorem we see that the limit is 0 although the function is not defined at the origin.

Earlier you answered two of my questions with yes. But can you prove it?

I wonder if there is a case where
1. Using polar coords evualtes a limit which is incorrect in that a
different value should be the correct one
'I can assure you this won't ever happen. :)'

2. Show that a limit does not exist where in fact there does exist a limit.
'This won't happen, either. :)'

I will try to informally prove it.
we are able to convert x and y into polar form x=rcos(theta), y=rsin(theta)

so f(x,y)=g(r,theta)
we want theta to vary over the domain of the real numbers so that g is not just a point on the plane but a circular disk forming a neighborhood around some origin. We let h(theta)=theta, theta is a subset of the reals.

so f(x,y)=g(r,h(theta))

If we want to know the limit of f as (x,y)->(0,0), it would be equivalent to the limit of g as (r,h(theta))->(0,theta). where theta is part of the reals.

In most cases we can simply evaluate the limit of g by letting r=0. We might get a defined single value where theta and r dissapear. We know in this case that the a definite limit exists for g hence it is also the limit for f since f=g.

If the limit of g is evaluated to be defined but multivalued such as having terms with theta than no definite limit exist for g hence none will exist for f since f=g.

If the limit of g is evaluated to be undefined than we will have to resort to other means instead of simply let r=0. For example use the sandwich rule or standard limits which is what you have done with some of your examples. If none of these methods work than we are pretty sure that a limit does not exist.One suspician I have with using polar coords is that there might exist some complicated function which have a limit but can only be known by a very complicated and weird path into the origin. The polar coords is not able to pick it up and nor are the other tools like the sandwich theorem be useful.

 

[VietDao29]

'A point z is a limit point of a set S (real or complex) if any neighbourhood of z contains a point of S, other than z itself.'
Hence only one point (meaning one path into the origin) is enough to prove that a limit exists.
So with your example, even if the function is undefined going along the x or y axis, it does not mean that another path cannot approach the origin legitamatly. Therefore there might be a point around the neighbourhood of the origin that is an element of the function $$\lim_{(x, \ y) \rightarrow (0, \ 0)} |xy| \ln|xy|$$

Arghhh,... uhmmm, I think you should read my post carefully.
I take that example from this thread. Let's see if you can get it. The limit does not exist, not 0.
This limit:
$$\lim_{(x, \ y) \rightarrow (0, \ 0)} (|x| + |y|) \ln(|x| + |y|) = 0$$
Can you see the difference?

One suspician I have with using polar coords is that there might exist some complicated function which have a limit but can only be known by a very complicated and weird path into the origin. The polar coords is not able to pick it up and nor are the other tools like the sandwich theorem be useful.

As I told you before, you should be flexible. Some function like sine, and cosine, logarithm, blah, blah, blah... then you should use something else, polar coord won't help these cases, I think.
Now look at the thread I've shown you to see if you can get it. :)

 

[pivoxa15]

Uhmm, this is not quite correct... :)

3. There is no neighbourhood of (0, 0) that the function is defined for every x, and y in that neighbourhood.
------------------
The limit:
$$\lim_{(x, \ y) \rightarrow (0, \ 0)} |xy| \ln|xy|$$
does not exist. Since the function is not defined along the x, or y-axis (number 3).
The limit still does not exist even if it does evaluate to 0 as you approach it along the x = ky path, or y = kx (k is not 0), or even in a parabolic path like y = x² (Or whatever path you like )...
However, the limit $$\lim_{(x, \ y) \rightarrow (0, \ 0)} (|x| + |y|) \ln (|x| + |y|) = 0$$ (explained in the previous post). Notice that the function is defined for every pair of (x, y), but (0, 0). Right?

How sure are you have criteria 3? Where did you get the information from?

Why do you need a whole defined (by the function) neibourhood around the origin in order for the limit to exist? Why not just allow at least one point on any neighbourhood around the origin to exist?

 

[VietDao29]

How sure are you have criteria 3? Where did you get the information from?

Why do you need a whole defined (by the function) neibourhood around the origin in order for the limit to exist? Why not just allow at least one point on any neighbourhood around the origin to exist?

By the definition of limit of 2 variable function, we haveL
$$\lim_{\substack{x \rightarrow p \\ y \rightarrow q}} f(x, \ y) = L$$
if and only if:
For every ε > 0 there exists a δ > 0 such that for all real numbers x, y with 0 < ||(x,y)-(p,q)|| < δ, we have |f(x,y)-L| < ε
Where ||(x,y)-(p,q)|| represents the Euclidean distance.
Now, let's look at the bolded part:
0 < ||(x,y)-(p,q)|| < δ, that means that there must exist a circular region with radius δ (excluding the center (p, q)), such that for every point (x, y) in that region, we have: |f(x,y)-L| < ε.
So if there happens to be some point (x, y) in that circular region that the function f(x, y) is not defined, then is |f(x,y)-L| defined?
No, right? So can we say that: for every number x, y in that region, the following statement is true |f(x,y)-L| < ε? Again, it's no, right?
So again, if there exists no circular region arround (p, q) (excluding the point (p, q) itself), that for every pair of number (x, y) the function f(x, y) is defined, then the limit does not exist (as it violates the definition).
Now can you get it? :)

 

[pivoxa15]

I get it for this example. Can you prove why polar coords work and why sometimes it does not work?

 

[VietDao29]

I get it for this example. Can you prove why polar coords work and why sometimes it does not work?

It will always work. But as I said earlier, sometimes it will take more time to solve a problem in polar coord, than in some way else. Consider this example:
$$\lim_{(x , \ y) \rightarrow (0, \ 0)} \frac{1 + x ^ 2 + y ^ 2}{y ^ 2} (1 - \cos y)$$
Now since we have:
$$\lim_{y \rightarrow 0} \frac{1 - \cos y}{y ^ 2} = \frac{1}{2}$$. The whole expression will tend to:
$$\lim_{(x , \ y) \rightarrow (0, \ 0)} \frac{1 + x ^ 2 + y ^ 2}{y ^ 2} (1 - \cos y) = \frac{1}{2} (1 + 0 ^ 2 + 0 ^ 2) = \frac{1}{2}$$.
But if you change to polar coordinate:
$$\lim_{(x , \ y) \rightarrow (0, \ 0)} \frac{1 + x ^ 2 + y ^ 2}{y ^ 2} (1 - \cos y) = \lim_{r \rightarrow 0} \frac{1 + r ^ 2}{r ^ 2 \sin ^ 2 \theta} (1 - \cos (r \sin \theta))$$. This will take more time to solve, and look much more complicated than the previous way I've shown you right?
So overall, if there are some functions like sine, cosine, tangent, cotangent, exponential, blah blah blah, don't use polar-coord.
Only use polar coordinate when you encounter some fraction, of which both numerator, and denominator are polynomials.
Can you get this? :)