Let $a \in ( -\infty, \infty)$. Suppose $\lim_{x \rightarrow a} f(x) = L \neq0$. Use the $\epsilon-\delta$ argument to prove $$\lim_{x\rightarrow a} \frac{1}{f(x)}= \frac{1}{L}.$$
I know proof should include below but i cannot proceed
$$0<|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon$$
$$\left|\frac{1}{f(x)}-\frac{1}{L}\right| =\frac{\left|L-f(x)\right|}{\left|Lf(x)\right|}.$$
From where you are now, you know you can take $|x-a|$ small enough to make the numerator arbitrarily small. The problem arises if the denominator is too small. The trick is that because $f(x)$ is very close to $L$, we can take $|x-a|$ small enough that $|f(x)|>|L|/2$, by finding $\delta$ which corresponds to $\varepsilon=|L|/2$. If we do, then
$$\frac{|f(x)-L|}{|L f(x)|} \leq \frac{2}{L^2} |f(x)-L|$$
Now you just have to manage that constant in front.
