# Limit of a two variable function

• I
I am not saying I agree with the answer, just that I have never tried that approach before. One problem with letting x=rcos(t) and y=rsin(t) is that it limits curve to travel at any point on the unit circle (assuming r=1 for arguments sake) for any value of t, which supports your comment about this problem's ability to have a limit that exists. Since we do not know where the problem came from (text book, etc). It could be that limit does not exist as stated and the sources answer is wrong to begin with. My next thought on this problem is to rewrite the numerator using identities and break it up into two separate limits and see where that goes.

haushofer
I'm trying to verify that: $$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x+y^{2}}=0.$$
$$0<\sqrt{x^2+y^2}<\delta\rightarrow |\frac{\sin (x^{3}+y^{3})}{x+y^{2}}|<\epsilon$$
$$0\leq |\sin (x^{3}+y^{3})|\leq |(x^{3}+y^{3})|\leq |x|x^2+|y|y^2$$
$$|\frac{\sin (x^{3}+y^{3})}{x+y^{2}}|\leq \frac{ |x|x^2+|y|y^2}{|x+y^{2}|}$$
So I'm stuck here because of the denominator |x+y²|. What can I do?
My 2 cents:

I'd evaluate the limit along a path like y=kx for real k; if the limit does not depend on k, it exists, and you can try to manage it in a form to apply the squeeze theorem. So, evaluating the limit along the path y(x)=kx gives
$$\lim_{ x\rightarrow 0}\frac{\sin ([k^3+1] x^3)}{x+k^2 x^2}$$

Now use the inequality
$$-t \leq \sin(t) \leq +t$$

for real t to constrain the limit by

$$- \frac{[k^3+1] x^3}{x+k^2 x^2} \leq \lim_{ x\rightarrow 0}\frac{\sin ([k^3+1] x^3)}{x+k^2 x^2} \leq \frac{[k^3+1] x^3}{x+k^2 x^2}$$

Using L'Hospital on the left and right gives

$$\lim_{x\rightarrow 0} \pm \frac{[k^3+1] x^3}{x+k^2 x^2} = \lim_{x\rightarrow 0} \pm \frac{[k^3+1] 3x^2}{1+2k^2 x} = \pm \frac{0}{1+0} = 0$$

so according to the squeeze theorem one has

$$\lim_{ x\rightarrow 0}\frac{\sin ([k^3+1] x^3)}{x+k^2 x^2} =0$$

Since this is true for all k, the answer depends not on the angle to which one arrives at the point (x,y)=(0,0), so the limit exist
and is equal to

$$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x+y^{2}}=0$$

That makes complete sense when using the squeeze theorem for the path y(x) =kx. Earlier in the posts where y=x was tried it, would have worked then if the squeeze theorem was applied at that point in time to the problem. I forgot all about suggesting that theorem be tried again. Thank you!

haushofer
Trying the path y(x)=x can only indicate that the limit does not exist if it is supplemented by another path giving different answers. That's the nice thing about paths like y(x)=kx or switching to polar coordinates or other curve representations that cover all the directs in which one approaches the limit point (x,y); they are general.

Stephen Tashi
Since this is true for all k, the answer depends not on the angle to which one arrives at the point (x,y)=(0,0), so the limit exist

That is not a valid conclusion. The existence of a common value for the limit along a family of straight line paths does not imply that the limit as (x,y) -> (0,0) exists. For example, see the answer to http://math.stackexchange.com/quest...existence-of-limit-of-a-two-variable-function

Stephen Tashi
This limit does not exist. Note that the function isn't even defined on the path x=-y^2,
That's correct, but does that, by itself, prevent the limit from existing? This is very technical question involving how we interpret the definition of the limit. The definition has the form " For each ... there exists...such that...for each ##(x,y)##, ##| f(x,y) - L| < \epsilon##". ##\ ## Are we to understand that the implicit meaning is "for each ##(x,y)## in the domain of ##f##"? ##\ ## Or perhaps that ##| f(x,y) - L | < \epsilon##" is not a false statement when ##(x,y)## is not in the domain of ##f##, but rather an undefined expression?

and if you consider paths close to that path (e.g. x = -y^2 + y^4), the limit diverges to infinity.
That's a good idea, but does it work? ##\ | sin(\theta) | \le |\theta | ##. ##\ \theta## would be a polynomial of degree 12 in ##y## and the denominator would be a polynomial of degree 4 in ##y##.

Stephen Tashi
I'm trying to verify that: $$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x+y^{2}}=0.$$

One thought is that ##|\sin(\theta)| \le |\theta| ##. ##\ ## Argue that for ##(x,y) \ne (0,0)##, $$| \frac{ \sin(x^3 + y^3)}{x + y^2} | \le |\frac{x^3 + y^3}{x + y^2}|$$. Then try polar coordinates on the ratio of the polynomials.

$$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x^{3}+y^{3}}\frac{x^{3}+y^{3}}{x+y^{2}}=\lim_{(x,y)\rightarrow (0,0)} \frac{x^{3}+y^{3}}{x+y^2}$$

That's a good intuitive idea, but notice how hard it is to make that algebra into correct logic. For example, it assumes at the outset that ##lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x^{3}+y^{3}}\frac{x^{3}+y^{3}}{x+y^{2}}## exists. Of course, some instructors (and many home work graders) prefer to see algebra instead of logic.

The basic pattern that obeys logic would be:
Show ##lim_{(x,y) \rightarrow (0,0)} p(x,y) = L_1##
Show ##lim_{(xy) \rightarrow (0,0)} q(x,y) = L_2##
Conclude that ##lim_{(x,y) \rightarrow (0,0)} p(x,y)q(x,y) = L_1 L_2##.

Polar coordinates might be a good way to tackle ## \lim_{(x,y)\rightarrow (0,0)} \frac{x^{3}+y^{3}}{x+y^2}## instead of using further algebra with ##x##'s and ##y##'s.

How about this approach. I plotted this function using quick graph. There are no breaks in the graph at the origin so therefore it is continuous there. Also the numerator sin(x^3 + y^3) can be rewritten as sin(x^3)cos(y^3) + cos(x^3)sin(y^3). Now that the numerator has been rewritten as a sum of the products of periodic continuous functions it should make it easier to prove the existence of the limit. Even though the denominator is undefined at (0,0), the numerator is approaching 0 faster than the denominator in terms of how the graph is changing.

Stephen Tashi
There are no breaks in the graph at the origin so therefore it is continuous there.

The function ##\frac{\sin(x^3 + y^3)}{x + y^2}## is not continuous at ##(0,0)##. Have you studied the definition of "##f(x,y) ## is continuous at ##(a,b)##" ?

i don't understand the "*. *". Comment. This site is not about PROVING someone is wrong or putting people down which you did with me. The dialog in these forums is not supposed to be judgmental towards others. So tell me why you disagree without being judgmental. What do you see in detail is wrong with my approach ?

See my note

That's correct, but does that, by itself, prevent the limit from existing? This is very technical question involving how we interpret the definition of the limit. The definition has the form " For each ... there exists...such that...for each ##(x,y)##, ##| f(x,y) - L| < \epsilon##". ##\ ## Are we to understand that the implicit meaning is "for each ##(x,y)## in the domain of ##f##"? ##\ ## Or perhaps that ##| f(x,y) - L | < \epsilon##" is not a false statement when ##(x,y)## is not in the domain of ##f##, but rather an undefined expression?

All very true, but actually orthogonal to the point I was trying to make. What I was trying to get at is that if you are trying to find the limit of a rational function (or something which is effectively a rational function, since $\sin (\theta) ~ \theta$ for small $\theta$) at the origin, the denominator being zero along some curve running through the origin usually means that you can make the function blow up by following another curve close to the first one. Thus seeing that the function fails to exist along some curve through the origin is usually a good sign that the limit won't exist, independently of the answer to that technical question.

That's a good idea, but does it work? ##\ | sin(\theta) | \le |\theta | ##. ##\ \theta## would be a polynomial of degree 12 in ##y## and the denominator would be a polynomial of degree 4 in ##y##.

Yes. Along the curve x=-y^2 + y^4:

$$\lim_{y \to 0^+} \frac{\sin (x^3 + y^3)}{x + y^2} \\ = \lim_{y \to 0^+}\frac{\sin ((-y^2 + y^4)^3 + y^3)}{-y^2 + y^4 + y^2} \\ = \lim_{y \to 0^+}\frac{\sin (y^3 - y^6 + 3y^8 - 3y^{10} + y^{12})}{y^4} \\ = \lim_{y \to 0^+}\frac{y^3 + O(y^6)}{y^4} \\ = \lim_{y \to 0^+}\frac{1 + O(y^3)}{y} \\ = \infty$$

So the original limit does not exist.

• Stephen Tashi and Inventive
All very true, but actually orthogonal to the point I was trying to make. What I was trying to get at is that if you are trying to find the limit of a rational function (or something which is effectively a rational function, since $\sin (\theta) ~ \theta$ for small $\theta$) at the origin, the denominator being zero along some curve running through the origin usually means that you can make the function blow up by following another curve close to the first one. Thus seeing that the function fails to exist along some curve through the origin is usually a good sign that the limit won't exist, independently of the answer to that technical question.

Yes. Along the curve x=-y^2 + y^4:

$$\lim_{y \to 0^+} \frac{\sin (x^3 + y^3)}{x + y^2} \\ = \lim_{y \to 0^+}\frac{\sin ((-y^2 + y^4)^3 + y^3)}{-y^2 + y^4 + y^2} \\ = \lim_{y \to 0^+}\frac{\sin (y^3 - y^6 + 3y^8 - 3y^{10} + y^{12})}{y^4} \\ = \lim_{y \to 0^+}\frac{y^3 + O(y^6)}{y^4} \\ = \lim_{y \to 0^+}\frac{1 + O(y^3)}{y} \\ = \infty$$

So the original limit does not exist.
Thanks for taking the time to explain it this way. I can visualize what you are saying

Stephen Tashi
Yes. Along the curve x=-y^2 + y^4:

I see what you mean.

We can also use the way more often seen in textbooks - showing that different limits can be had by using different curves if we use curves of the form ##x = -y^2 + ky^3##.

Let ##G(x,y) = \frac{x^3 + y^3}{x + y^2}\ ## Use the path defined by ##x = -y^2 + k y^3## where ##k## is a non-zero constant. This path goes through ##(0,0)##

Along this path, for ##y \ne 0##, ##G(x,y) = g(y) = \frac{ (-y^2 + ky^3)^3 + y^3}{ky^3} = \frac{(y^6 + 3ky^7 - 3k^2 y^8 + k^3 y^9 ) + y^3}{ky^3} = \frac{y^3+ 3ky^4 - 3k^2y^5 + y^6 + 1}{k} ##

So ##lim_{y \rightarrow 0} g(y) = \frac{1}{k}##.

Since ##\sin()## is a continuous function ##lim_{y \rightarrow 0} \sin(g(y)) = \sin( \frac{1}{k})##.

This shows ##lim_{(x,y) \rightarrow (0,0)} sin(G(x,y)) ## does not exist since we have can get different limits on different paths depending on our choice for ##k##.

Thank you Stephen for the further information

haushofer