Originally Posted by
Scott H Is $\displaystyle p$ continuous?
Since $\displaystyle p(x)^2$ doesn't distinguish between the sign of $\displaystyle p(x)$, we may assume that $\displaystyle p(x)$ is nonnegative everywhere. Now, let's suppose that $\displaystyle c$ is a value at which $\displaystyle p(c)=L\ne 0$. By the assumed continuity of $\displaystyle p$, we have, from the epsilon-delta definition of limit,
For all $\displaystyle \epsilon$, there exists a $\displaystyle \delta$ such that $\displaystyle 0<|x-c|<\delta$ implies $\displaystyle |f(x)-L|<\epsilon$.
This just states that $\displaystyle \lim_{\small x\rightarrow c}f(x)=f(c)$.
What we need to do is find a way to show that there is some positive area under the curve $\displaystyle y=p(x)$. What happens when we let $\displaystyle \epsilon=\frac{L}{2}$?