Theorem: If $\lim_{x\to a}\ g(x)=M\ne 0$ then $\lim_{x\to a}\ \frac{1}{g(x)}=\frac{1}{M}$.
Proof:

Suppose $\lim_{x\to a}\ g(x)=M\ne 0$, and consider any $\epsilon >0$. We must find a $\delta>0$ such that if $0< |x-a|< \delta$ then $|\frac{1}{g(x)}-\frac{1}{M}|< \epsilon$.

First, observe that $|\frac{1}{g(x)}-\frac{1}{M}|=\frac{|M-g(x)|}{|Mg(x)|}$. The numerator can be made small, but we also have to show that the denominator is not small when $x$ is near $a$.

Since $\lim_{x\to a}\ g(x)=M\ne 0$ we can find a $\delta_1>0$ such that if $0<|x-a|< \delta_1$ then $|g(x)-M|< \frac{|M|}{2}$.

Therefore $|M|=|M-g(x)+g(x)|\le|M-g(x)|+|g(x)|< \frac{|M|}{2}+|g(x)|$.

It follows that $|g(x)|>\frac{M}{2}$, and so $\frac{1}{|Mg(x)|}=\frac{1}{|M||g(x)|}< \frac{1}{|M|}\cdot\frac{2}{|M|}= \frac{2}{M^2}$.

Also, there exists a $\delta_2>0$ such that if $0< |x-a|< \delta_2$ then $|g(x)-M|< \epsilon(\frac{M^2}{2})$.

Take $\delta=\text{min}\{\delta_1,\delta_2\}$ and assume $0< |x-a|< \delta$.

Then $0< |x-a|< \delta_1$ and $0<|x-a|< \delta_2$, so $|\frac{1}{g(x)}-\frac{1}{M}|=\frac{|M-g(x)|}{|Mg(x)|}< (\frac{2}{M^2})\cdot\epsilon\cdot(\frac{M^2}{2})=\epsilon$.

This proves that $\lim_{x\to a}\ \frac{1}{g(x)}=\frac{1}{M}$.