# Show that limit does not exist (two variables)

I just started looking into multiple variable calculus and limits involving them. I'm not amazing at limits either.

I want to answer this question:

Show that the following limit does not exist

$$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2y^2+(x-y)^2}$$

So, my working:

$$\lim_{x\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}=\lim_{x\to 0}\frac{0}{y^2}=0$$

and

$$\lim_{y\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}=\lim_{y\to 0}\frac{0}{x^2}=0$$

I wasn't sure what to do after this as they're both 0 but using the fact it can approach from any direction, I tried substituing $$y=x$$, not sure if that's correct - or my working.

So, let $$y=x$$, then

$$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2x^2+(x-y)^2}=\lim_{x\to 0}\frac{x^4}{x^4+(x-x)^2}=1$$

Therefore limit doesn't exist.

Is this somewhat correct? What's the best way to answer a question like this?

Also, when showing that this limit does not exist, do I need to find different values of limits for both $${x\to 0}$$ AND $${y\to 0}$$? Or is one enough, e.g. if I just find two different values for two limits for $${x\to 0}$$ without using $${y\to 0}$$ in my calculation at all, is that fine?

Thanks!

What you have done is correct. The limit exists only if the value of the limit along every direction that leads to $$(0,0)$$ is same.

So when you calculate $$\lim_{x\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}$$ you are calculating limit along the line $$x=0$$.

Similarly,
$$\lim_{y\to 0}\frac{x^2y^2}{x^2y^2+(x-y)^2}$$ is limit along line $$y=0$$.

And the last limit you calculated is along line $$y=x$$.

So to answer your question, yes it would have been perfectly acceptable if you did not calculate limit along $$y=0$$. Just showing two examples where the limit comes out to be different along different directions is enough to show limit does not exist.

• "The limit exists only if the value of the limit along every direction ... is same" A word of caution: that's a necessary, but not a sufficient condition. It can happen that all directional limits give the same value, and the limit does not exist. – leonbloy Feb 6 at 14:35
• @leonbloy Do you have an example of that case? – Shufflepants Feb 6 at 15:48
• @leonbloy I don't know of any way by which we can prove all directional limits exist and have same value, except by using epsilon-delta definition. Though I have very limited knowledge of multi-variable calculus. So it would be great if you could give an example where all directional limits give the same value, and the limit does not exist. – Swapnil Rustagi Feb 6 at 15:53
• If all directional limits give the same value, then the function might not be continuous at that point; it would need to be defined at that point, with a value equal to all the limits. But I think if all possible directional limits exist and are equal, then the function can be said to have that limit. (Unless anyone has a counterexample?) – gidds Feb 6 at 16:28
• @Shufflepants $f(x,y)=xy^2/(x^2+y^4)$ See here ocw.umb.edu/mathematics/math-240-calculus-iii-spring-2010/… (from page 19) – leonbloy Feb 6 at 17:11

Your reasoning is ok. If the limit existed, you would obtain the same value for every directional limit, which was not the case.

• How can I improve my reasoning? – Mandingo Feb 6 at 11:48
• No much room for improvement here... It is pretty standard. Maybe you could have just used directional limits instead of starting with the iterated limits, but in the end it is quite the same. – PierreCarre Feb 6 at 11:54

What you did is completely correct, but perhaps you can ease it a little bit as follows:

For $$\;x=0\;,\;\;y\to0\;$$ the limit clearly is $$\;0\;$$, whereas for $$\;x=y\;$$ and $$\;x\to0\;$$ we get

$$\frac{x^4}{x^4+(x-x)^2}=1\xrightarrow[x=y\to0]{}1$$

Thus the limit depends on the path chosen $$\;\implies\;$$ the limit doesn't exist...and this is the relevant argument: going on different paths yields different limits.

You did right.

A possible improvement is to compute the limit along an arbitrary line: along $$y=mx$$ you have to compute $$\lim_{x\to0}\frac{x^2(mx)^2}{x^2(mx)^2+(x-mx)^2}= \lim_{x\to0}\frac{m^2x^4}{m^2x^4+x^2(1-m)^2}= \lim_{x\to0}\frac{m^2}{m^2+(1-m)^2x^{-2}}= \begin{cases} 0 & m\ne 1 \\[4px] 1 & m=1 \end{cases}$$ Doing this way may immediately show how to solve the business. Of course, if you find that all limits along lines are equal, you cannot conclude that the limit exists; instead, you should try along other curves, if you suspect the limit doesn't exist.

What you did is correct, but note that after proving that the limit is $$0$$ when you take $$y=0$$, the fact that it is also $$0$$ when you take $$x=0$$ is irrelevant. What matters here is that going to $$(0,0)$$ through two different directions leads you to two distinct limits. Therefore, there is no (global) limit at $$(0,0)$$.