Determining if a limit does not exist.

Question

I am having trouble determining if the limit defined below exists. I do not believe it does, but I am not sure how to find 2 paths where the limits differ. I've tried $f(x, mx)$, $f(my, y)$, $f(x, mx^2)$, but nothing really seems to work.

Show that the limit does not exist:

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

I thought about parameterizing using polar coordinates, but I wasn't sure how to show that the $\displaystyle \lim_{r^+ \to 0}$ will depend on $\theta$ (and hence does not exist).

Any help appreciated. Thanks.

If I set y = 0, then you are taking $lim_{(x,0) \to (0,0)}$, right? — TimelordViktorious, Apr 14 '17 at 18:21
Thank you. I guess I overcomplicated this question by approaching via paths. — TimelordViktorious, Apr 14 '17 at 18:22
You can apply L'Hopital's rule for y = 0, but it gets messy. It doesn't look like squeeze theorem is helpful either. — TimelordViktorious, Apr 14 '17 at 18:29

score 1 · Accepted Answer · answered Apr 14 '17 at 18:40

Since $1-\cos(x) =2\sin^2(x/2) $, since $\sin x \approx x$ as $x \to 0$, $\frac{y^2 + (1-\cos(x))^2}{x^4 + y^2} =\frac{y^2 + 4\sin^4(x/2)}{x^4 + y^2} \approx \frac{y^2 + x^4/4}{x^4 + y^2} $.

We can now apply a standard technique and let $y = k x^2$. This becomes $\frac{k^2x^4 + x^4/4}{x^4 + k^2x^4} =\frac{k^2 + 1/4}{k^2+1} $ which will have different values for different values of $k$.

Determining if a limit does not exist.

1 Answers1