Limit Proof Help I could use some help on a proof. Suppose $f(x)≥0$ for all $x$ (except perhaps at $x=a$). Show that if $\lim\limits_{x→a}f(x)=L$, then $L≥0$. Any guidance would be helpful.
 A: Suppose not. Then we can find $\delta $ such that $|f(x) - L| = f(x) - L < - L$ for $0 <|x-a | <\delta $ - that is, $f(x) <0,$ but this is absurd. 
A: Assume that $L < 0$. Then there is an $\delta > 0$ such that if $\lvert f(x) - L\lvert < \frac{-L}{2}$ whenever $\lvert x - a\lvert < \delta$. Now consider:
$$
\begin{align}
\lvert f(x) - L\lvert &< \frac{-L}{2} & \Rightarrow \\
\frac{L}{2} < f(x) - L &< \frac{-L}{2} & \Rightarrow \\
\frac{3L}{2} < f(x) &< \frac{L}{2}.
\end{align}
$$
So for all $x$ for which $\lvert x - a\lvert < \delta$ you have
$$
\frac{3L}{2} < f(x) < \frac{L}{2}.
$$
Can you finish the argument from here?

Edit: To add a bit more explanation here. What we am doing above is trying to prove the statement by assuming the the negation of the statement. I.e. we assume that what we want to prove is false. Then we want to prove that the assumption is false. That is why we start out by assuming that $L<0$. What we want to prove is that there is some $x$ such that $f(x) < 0$. Above then we get to the equation that for $x$ in the interval $(x-a, x+a)\setminus \{a\}$ we have
$$
\frac{3L}{2} < f(x) < \frac{L}{2} <0.
$$
But that tells us exactly that for any $x \in  (x-a, x+a)\setminus \{a\}$, $f(x) < 0$.
We say that we have proved the statement by contradiction.
