Can squeeze theorem be used to prove the nonexistence of a limit? For squeeze theorem, if $0 \leqslant f(x) \leqslant 1$, can this be used to prove the assertion that the limit $f(x)$ does not exist? I.e. $f(x)$ is $xy/ (x^2+xy+y^2)$, as $(x,y)\rightarrow(0,0)$.
I'm aware you can use the method of approaching from different paths, but was wondering if squeeze theorem was enough? My first guess is obviously this isn't enough to prove anything, since $f(x)$ can still be $0$, but of all the examples I tried, when the squeeze theorem doesn't squeeze $f(x)$ into a number $L$, the limit doesn't exist. Just a coincidence?
 A: As I know the squeeze theorem, it says if (blah) then the limit is $L$.  It says nothing if you can't prove (blah).  Maybe there is a better thing to squeeze it, maybe there is no limit.
A: It is probably a coincidence. However, you should be careful when the squeeze theorem apparently works.
Consider the function
$$f(x,y) = \frac{x^2 y}{x^4 + y^2}.$$
We want to find its limit at the origin. Let's switch to polar coordinates by making $x = r \cos \theta$ and $y = r \sin \theta$. The function takes the form
$$f(x,y) = f(r,\theta) = \frac{(r \cos \theta)^2 (r \sin \theta)}{(r \cos \theta)^4 + (r \sin \theta)^2} = \frac{r^3 \cos^2 \theta \sin \theta}{r^4 \cos^4 \theta + r^2 \sin^2 \theta} = \frac{r \cos^2 \theta \sin \theta}{r^2 \cos^2 \theta + \sin^2 \theta}.$$
I've made a little abuse of notation when I wrote $f(x,y) = f(r, \theta)$. You could try saying that
$$0 \leq \left\vert \frac{r \cos^2 \theta \sin \theta}{r^2 \cos^4 \theta + \sin^2 \theta} \right\vert \leq \frac{r}{r^2 \cos^4 \theta + \sin^2 \theta}$$
and conclude that the original limit is zero. The problem is that you would arrive at a false result. To see this, consider the path $y=x^2$. This gives us
$$f(x,x^2) = \frac{x^2 (x^2)}{x^4 +(x^2)^2} = \frac{x^4}{2x^4} = \frac{1}{2},$$
which is a path passing through the origin but where the function takes a different limit than zero, therefore the function cannot be continuous at that point.
What should sound an alarm is the fact that when we switched to polar coordinates the numerator had a $\sin \theta$ lying around. Depending on which angle we approached the origin, it could be negative. That information gets lost when we use the absolute value, leading to a wrong conclusion. Moreover, polar coordinates cover straight lines through the origin only, when a nonlinear path (such as $y=x^2$) could reveal the problem.
A: The existing answers are perfectly correct, but since this has gotten bumped: There is something like a converse of the squeeze theorem in the following sense.
If $f$ is a real-valued function on a metric space $(X, d)$, and if $a$ is a limit point of $X$, then $\lim\limits_{x \to a} f(x)$ exists and is equal to $L$ if and only if
\begin{align*}
\limsup_{x \to a} f(x)
  &:= \lim_{\delta \to 0}\; \sup_{0 < d(a, x) < \delta} f(x), \\
\liminf_{x \to a} f(x)
  &:= \lim_{\delta \to 0}\; \inf_{0 < d(a, x) < \delta} f(x)
\end{align*}
(which always exist as extended real numbers) are both equal to $L$.
Presumably when you found "$0 \leq f(x) \leq 1$" in your examples, $0$ and $1$ were actually the $\liminf$ and $\limsup$ of your function, i.e., the "worst cases". Since $0 \neq 1$, the function $f$ had no limit.
