Proving that $(\lim_{x \to a}f(x) = L) \Rightarrow (\lim_{x \to a}\frac{1}{f(x)} = \frac{1}{L})$ using Epsilon-Delta Using the Epsilon-Delta definition of a limit, how can I rigorously prove that permitting $f(a) \neq 0$, 
$$\left(\lim_{x \to a}f(x) = L\right) \Rightarrow \left(\lim_{x \to a}\frac{1}{f(x)} = \frac{1}{L}\right)$$
I tried to work backwards by simplifying: 
$$\left|\frac{1}{f(x)}-\frac{1}{L} \right| = \left|\frac{ L - f(x)}{Lf(x)} \right| = \left| f(x) - L \right|\frac{1}{|Lf(x)|}$$
but I am not sure how to proceed from here or if this is even useful to the overall proof as a whole.
 A: You are to prove that $\lim_{x \to a} \frac{1}{f(x)} = \frac 1L$. First of all, let $\epsilon > 0$.
We note that $\left|\frac{1}{f(x)} - \frac 1L\right| = \left|\frac{L - f(x)}{Lf(x)}\right| = \frac{|f(x) - L|}{|Lf(x)|}$.
Fix an $\epsilon_0 > 0$, that we will choose later. Let $\delta > 0$ be such that $|x - a| < \delta \implies |f(x) - L| < \epsilon_0$. Then, we see that $ L - \epsilon_0 < f(x) < L+ \epsilon_0$. Thus, if $\epsilon_0$ is chosen small enough (say $\left|\frac L2\right|$), then $f(x)$ cannot be zero if $|x - a| < \delta$, and will have the same sign as $L$. Then, we also see that $ L(L - \epsilon_0) <Lf(x) < L(L+\epsilon_0)$, where we see that all these expressions have the same sign, which is positive.
This implies : 
$$
\frac{|f(x) - L|}{|Lf(x)|} \leq \frac{\epsilon_0}{L(L - \epsilon_0)}
$$
Now, choose $\epsilon_0$ such that $\frac{\epsilon_0}{L - \epsilon_0} = \epsilon$, in the starting. If that quantity is greater than $|\frac L2|$, then we can take $\min(\epsilon_0,|\frac L2|)$ as the desired  $\epsilon_0$ for which the $\delta$ that works for $\lim f(x) = L$ also works for $\epsilon$ and the limit you need to prove.
