# real analysis hw question

Since $f(x_0)\neq 0$ (I will assume that $f(x_0)>0$, but the other case is similar) let $f(x_0)=\alpha$. Now choose $\epsilon=\frac{\alpha}{2}$, and the appropriate $\delta$. Then for $|x-x_0|<\delta$ you have that $0.