$f:\mathbb{R}\to\mathbb{R}$, continuous, such that $xf(x)>0$ when $x\neq 0$. Show that $f(0)=0$

HINT: Since $xf(x)>0$ whenever $x\ne 0$, you know that $f(x)>0$ when $x>0$, and $f(x)<0$ when $x<0$. (Why?) What does this say about $\lim\limits_{x\to 0^-}f(x)$ and $\lim\limits_{x\to 0^+}f(x)$?

$xf(x) > 0$ implies that $f(x) > 0$ when $x$ is positive and $f(x) < 0$ when $x$ is negative.

By intermediate value theorem, we must have $f(c) = 0$ at some value between a negative and positive number, since we know it is non-zero everywhere else it must occur at zero.

Since $f$ is continuous, if $f(0)=a\gt0$, then there is an $\epsilon\gt0$ so that for $|x|\le\epsilon$ $$ f(x)\ge\frac a2 $$ Then for $-\epsilon\le x\lt0$, $$ xf(x)\lt0 $$ which contradicts the hypothesis.

If $f(0)=a\lt0$, then there is an $\epsilon\gt0$ so that for $|x|\le\epsilon$ $$ f(x)\le\frac a2 $$ Then for $0\lt x\le\epsilon$, $$ xf(x)\lt0 $$ which contradicts the hypothesis.

Thus, we are left with the conclusion that $f(0)=0$.