Help with a multivariable limit involving absolute value

Question

I have been self-studying multivariable calculus and I need help with evaluating this limit - if it even exists.

Problem : $$\lim_{(x,y)\to (1,0)} \frac{|x-1|}{3y^\frac{2}{3}}$$

I think the limit might be 0, I am not sure, so far I have tried the problem like this :

$$\text{Set } f(x, y) = \frac{|x-1|}{3y^\frac{2}{3}} \text{ and suppose }$$ $$x=y+1 \implies f(y + 1, y) = \frac{|y|}{3y^\frac{2}{3}}$$ $$\text{ and as } x \to 1 \text{ we have } y \to 0 \text{.}$$ Now $$\lim_{y \to 0} f(y+1, y)$$ is 0, as $$\lim_{y \to 0+} f(y+1, y) = \lim_{y \to 0+} \frac{y^\frac{1}{3}}{3} = 0$$ $$\lim_{y \to 0-} f(y+1, y) = \lim_{y \to 0-} -\frac{y^\frac{1}{3}}{3} = 0$$

After this I don't know how to proceed, how can I generalise this? Should I try to do some $\epsilon-\delta$ proof showing that the limit is 0? I am not even sure if this limit even exists.

Answer 1

Well, the limit might be $0$. You have indeed shown that limit along the line $x=y+1$. But consider the path $x=1+y^{1/3}$? Or $x = 1+y^{2/3}$. Or $x=1+cy^{2/3}$ for different values of $c$? Then you get different answers. So the limit does not exist.

$\delta$-$\epsilon$ proofs are good for proving the limit exists, but usually paths chosen cleverly based on the actual function you have will give you intuition about what's going on. Just remember not to limit yourself to lines — pun intended.

Answer 2

When dealing with a two variables limit, we can naively say that if we can determine two paths along which the limits are different, then the limit does not exist. It's clear that we wouldn't ever be able to analyse the whole paths, since they are infinite. But there are some that often occurr. For example, the straight lines path. We consider the restriction along one of the infinite straight lines passing through the point $(x_0, y_0):

$$y = y_0 + m(x - x_0)$$

If the limit will depend upon $m$, then the limit does not exist, since a change in $m$ will result in a change in the value of the limit.

Now we choose the line passing through the point $(1, 0)$ so that $y = m(x-1)$

$$\lim_{x\to 1, y\to 0} \frac{|x-1|}{3y^{2/3}} \rightarrow \lim_{x\to 1} \frac{|x-1|}{3\sqrt[3]{(m(x-1))^2}} = \lim_{x\to 1} \frac{1}{3m^{2/3}}\cdot \frac{|x-1|}{\sqrt[3]{(x-1)^2}} = 0$$

In this case, and for this path, the limit does not depend upon $m$, and this means we cannot say anithing. Indeed a more intertwined path could exist, where the limit does depend upon $m$.

Well, for example let's analyse a simpler path: $y = m$ and $x = m$, that is along two lines parallel to the axis, one at a time.

When dealing along $y = m$ we have

$$\lim_{x\to 1} \frac{|x-1|}{3m^{2/3}} = 0$$

When running along $x = m$ we have

$$\lim_{y\to 0} \frac{|m-1|}{3y^{2/3}} = +\infty$$

Limits are different, hence the limit does not exist.