Hey guys, I just had a quick question regarding the squeeze theorem in multivariable calculus.

As an example, consider the function $f(x,y) = \frac{2xy}{x^2+y^2}$ and it's limit as $(x, y) \to (0, 0)$. Just out of interest, I attempted to apply the squeeze theorem:
$$-1 \lt \frac{x}{x^2+y^2} \lt 1 \\ -2 \lt \frac{2x}{x^2+y^2} \lt 2 \\ -2|y| \lt \frac{2xy}{x^2+y^2} \lt 2|y|$$

Since $\lim_{(x, y) \to (0, 0)} |y| = 0$, we can conclude (?) $\lim_{(x, y) \to (0, 0)} \frac{2xy}{x^2+y^2} = 0$. However, I'm pretty sure that this isn't correct as if we set $x = y$ and calculate the limit, we get a…

Let $x_1,...,x_n \in \mathbb{R}$ such that $f(x_1) = f(x_2) = \space ... \space = f(x_n)$
So we have the intervals $\left[x_1,x_2 \right],\left[x_2,x_3 \right],...,\left[x_{n-1},x_n \right]$
For each of those intervals, $f$ is either positive or negative.
We will call them positive intervals or negative intervals.
(1) We can prove that there are $\frac{n}{2}+1$ positive intervals or negative intervals.
Let say there are $\frac{n}{2}+1$ positive intervals. [other case is same]
By EVT,let $M_1,...,M_{\frac{n}{2}+1}$ by the maximums of each of those positive intervals.

Prithu: I suggest you should try for n=2 first to see that. On [0,1], let M be the maximum of f on [0,1] and suppose that it is attained at some c in [0,1]. So there exist a c' in R such that f(c)=f(c')=M. WLOG, let c'>c. On [c,c'], let min f[c,c'] be m then m<M and let this be attained at some a in (c,c') (note the open interval here).
By IVT, every value in [m=f(a), M=f(c')] should be attained by f on [a,c'] and the same set of values must be attained by f on [c,a].
But m must also be attained twice. What happens to f(x) when x>c' or when x<c?