Two Variable Limit: Is "0,0" Available?

[imana41]

please say is it this limit available on (0,0)

lim((x^2*y^2)/(x^2+y^2))

thanks

 

[Char. Limit]

What have you tried. Try taking the limit across x=0, y=0, and y=x, for the most common limits.

 

[imana41]

What have you tried. Try taking the limit across x=0, y=0, and y=x, for the most common limits.

all limits are 0 . is this limit exist on (0,0) ?
pleaes solve this for me with Definition of limit
thanks

 

[Kreizhn]

Whoa, are you sure all the limits are zero? You have to do some work yourself before we can help.

 

[Kreizhn]

Furthermore, the point of trying typical approaches is an attempt to show that the limit does not exist. If however all the approaches still give the same limit, it doesn't mean that there isn't a path that is inconsistent.

 

[imana41]

Whoa, are you sure all the limits are zero? You have to do some work yourself before we can help.

can you explain what is it's answer i can't Analysis this limit pleaes don't confuse me

 

[Kreizhn]

Well, if you try all the trajectories that Char. Limit gives, yes, they will all go to zero, but it would be nice to see that you've done some of the work there.

However, just because the limit from 3-paths are consistent doesn't mean that the limit exists. You need to apply stronger techniques. In particular, this problem is probably best handled by converting it to polar coordinates and examining the case when the radial coordinate tends to zero.

 

[imana41]

thanks kreizhn now i found
thanks

 

[Char. Limit]

Well, if you try all the trajectories that Char. Limit gives, yes, they will all go to zero, but it would be nice to see that you've done some of the work there.

However, just because the limit from 3-paths are consistent doesn't mean that the limit exists. You need to apply stronger techniques. In particular, this problem is probably best handled by converting it to polar coordinates and examining the case when the radial coordinate tends to zero.

My tests are more to see if a limit is inconsistent than if a limit is consistent, though.

 

[Kreizhn]

Certainly, and I wasn't trying to imply otherwise, so I'm sorry if it came across that way. However, in this case I don't believe those tests are very useful.

 

[imana41]

hi again
can you help me about this limits on (0,0)
1-lim((x*y)/(x^2+y^2))+(y*sin(1/x))
2-lim(x*sin(1/y)+y*sin(1/x))
3-lim((x+y)*sin(1/x)*sin(1/y))

 

[Metaleer]

imana41, I don't think you quite understand how this forum works. You're supposed to show us your work so that we know where you're stuck and so that we can give you a push in the right direction. We aren't homework help robots.

 

[imana41]

imana41, I don't think you quite understand how this forum works. You're supposed to show us your work so that we know where you're stuck and so that we can give you a push in the right direction. We aren't homework help robots.

your giude aren't very useful i don't find my first question compeletly they say this limit
lim((x^2*y^2)/(x^2+y^2)) aren't exist but i think it exist on (0,0)
and about other question i just wan't a useful help for solve two variable limit generaly. you confuse me Instead of push me.!

 

[Kreizhn]

Okay, but you have shown ZERO (0) attempt to do the questions yourself. Try to do them, and show us what you're doing. Then we'll give you the push. We're not going to solve the problem for you.

 

[Kreizhn]

And we didn't say that the limit doesn't exist. In all fairness, it might have seemed that way by what we were saying, but we were trying to get you to show us your work. I also think the limit exists, and suggested that you convert it to polar coordinates and do the limit from there.

 

[Kreizhn]

Hmmm...I'm not sure what that's supposed to say. Though I do see an r^2 \sin^2(\theta) which is a good sign. However, I think you're missing a cosine term.

Don't worry about using LaTeX if you're not sure how. We can decode the non-LaTeX stuff if necessary.

 

[Kreizhn]

Oh wait. Did you already simplify and use \cos^2(\theta) \sin^2(\theta) = \frac14 \sin^2(2\theta)? It makes more sense then, though this simplification was unnecessary. So what I think your LaTeX is supposed to say is

\lim_{r \to 0} \frac{ r^2\sin^2(2\theta) }4

and when you evaluate this limit you get?

 

[imana41]

zero

 

[Kreizhn]

Exactly. And the value of \theta doesn't matter, since the answer is zero regardless of what value of \theta you've chosen. Therefore, you're done! The limit is zero!

Now let's try another question.

 

[imana41]

about lim(ln(x*y^2)) on (0,0)
i think this is unavailable because ln(0)=undefined . is it true ?

 

[Kreizhn]

I think that the limit diverges, if that is what you mean. There is a difference between limits not existing, and going to infinity.

 

[imana41]

is the answer \infty

 

[Kreizhn]

Maybe -(negative) infinity?

 

[imana41]

Maybe -(negative) infinity?

yes i think so
and about lim(\frac{x^2*y^2}{x^3+y^3}) i can't any path for reject it but in polar system i
get attach i should show this equation hasne't limit on (0,0) but with what path

 

[Kreizhn]

Well, what happens if you take r \to 0 here? Do you not get the same result?

Edit: Granted, the path y=-x causes havoc, though I honestly can't remember if that affects our situation.

 

[imana41]

LaTeX Code: lim(\frac{x^2*y^2}{x^3+y^3}) if i get y=-x the limit unavailable on (0,0) and the limit undefined in zero is that mean the limit aren't exist in zero

 

[Kreizhn]

Yeah, I'm not sure how much help I'm going to be at this point. It's been far too long since I've had to do this stuff.

However, at this point I think that the polar coordinate definition is fine. You might want to make another thread for that particular question though. There's bound to be somebody around here with more expertise than me with this stuff.

Before you do that however, I would like to point out another way you could have solved the first question, as it may be of benefit to you later. You could use the squeeze theorem!

So consider
\lim_{(x,y) \to (0,0)} \frac{ x^2 y^2}{x^2 + y^2}
First of all, I claim that xy \leq \frac12 (x^2+ y^2). This follows from the fact that (x-y)^2 \geq 0 and
(x-y)^2 = x^2 + y^2 - 2xy \geq 0 \quad \Rightarrow xy \leq \frac12 (x^2 + y^2)
Then we notice that
0 \leq \frac{x^2 y^2}{x^2 + y^2} \leq \frac{xy(x^2+y^2)}{2(x^2+y^2)} = \frac12 xy
and using the squeeze theorem, the result follows.

 

[imana41]

thanks

 

[Metaleer]

One has to be careful with polar coordinates.

For instance, consider

f(x, y) = \frac{2 x^2 y}{x^4 + y^2}.​

We decide to switch to polar coordinates to study the behavior of this function:

f(r \cos \theta, r \sin \theta) = \frac{r \cos \theta \sin 2\theta}{r^2 \cos^4 \theta + \sin^2 \theta}.​

Holding \theta constant and taking the limit as r \rightarrow 0, we see that the limit is 0. Many people would think that since polar coordinates yields a particular value for the limit, this is it, and we have finished with the conclusion that this is the value of the limit.

Now, what happens when we decide to see what the limit is along y = x^2?:

f(x, x^2) = \frac{2 x^2 x^2}{x^4 + x^4},​

so,

\lim_{x \rightarrow 0}f(x, x^2) = \lim_{x \rightarrow 0} \frac{2 x^2 x^2}{x^4 + x^4} = 1​

How can this be? Polar coordinates says it's 0, and computing the limit along y = x^2 says it's 1!

The reason is that if by polar coordinates we obtain that the limit is 0, the original limit in rectangular coordinates is also 0 if and only if there exists a function F(r) that meets the two following requirements:

|f(r \cos \theta, r \sin \theta)| \leq F(r),~ \forall \theta​

and

\lim_{r \rightarrow 0}F(r) = 0​

In your previous example, there does exist such a function, because

\frac{1}{4}|r^2 \sin^2 2 \theta| \leq \frac{r^2}{4}​

and it just so happens that r^2/4 \rightarrow 0 as r \rightarrow 0. Notice however that in the counterexample I've presented, it is not possible to establish such a function.