Prove $\lim\limits_{x\to a} \frac{1}{g(x)} = \frac{1}{m}$ Suppose $\lim_{x\to a} g(x) = m$ and $m \neq 0$ then, $$\lim_{x\to a} \frac{1}{g(x)} = \frac{1}{m}$$
My attempt to prove this is using the epsilon-delta definition of a limit and here's what I have so far:
$\forall ε>0, \exists δ>0$, such that for all x, if $0<|x-a|<δ$ then $|g(x) - m| < ε$
$$|g(x) - m| < ε$$
I picked $δ = \frac{ε}{2}$ then,
$$|g(x) - m| < \frac{ε}{2} \Rightarrow m-\frac{ε}{2} < g(x) < \frac{ε}{2} + m \Rightarrow$$
$$\frac{1}{m-\frac{ε}{2}} < \frac{1}{g(x)} < \frac{1}{\frac{ε}{2} + m}$$
so,
$$|g(x) - m| < \frac{1}{g(x)} * \frac{1}{\frac{ε}{2} + m} < ε$$
$$δ = \frac{ε}{\frac{1}{\frac{ε}{2} + m}}$$
At this point I'm completely lost and have no idea how to proceed. Any help with what to do next would be appreciated.
 A: The proof went astray partly because $\epsilon$ was doing double-duty. 
We solve the problem for positive $m$. A similar argument will work for negative $m$. The two arguments can be combined.  
We want to show that for every $\epsilon \gt 0$, there is a $\delta$ such that 
$$\left|\frac{1}{g(x)}-\frac{1}{m}\right|\lt \epsilon$$
whenever $|x-a|\lt \delta$. Note that
$$\left|\frac{1}{g(x)}-\frac{1}{m}\right|=\frac{|g(x)-m|}{|g(x)|m}.$$
Now we proceed informally for this paragraph. We can certainly make $|g(x)-m|$ small. But we might have a problem if $|g(x)|$ is small. So we better make sure it isn't. Since $g(x)$ has limit $m$, after a while it is close to $m$, and therefore not tiny. Now back to formality. 
There is a $\delta_1$ such that if $|x-a|\lt \delta_1$, then $|g(x)-m|\lt \frac{m}{2}$. So if $|x-a|\lt \delta_1$, we have $g(x)\gt \frac{m}{2}$, and therefore 
$$\frac{|g(x)-m|}{|g(x)|m}\lt \frac{2}{m^2}|g(x)-g(a)|.\tag{1}$$
Now because $\lim_{x\to a} g(x)=m$, there is a $\delta_2$ such that if $|x-a|\lt \delta_2$, then $|g(x)-a|\lt \frac{m^2}{2}\epsilon$.
Let $\delta=\min(\delta_1,\delta_2)$. If $|x-a|\lt \delta$, then by (1) we have $\left|\frac{1}{g(x)}-\frac{1}{m}\right|\lt \epsilon$. 
A: I think this would be neater.
$0<|x-a|<\delta \Rightarrow |g(x)-m|<\varepsilon$
Taking $\varepsilon = \frac{|m|}{2}$ there is a $\delta^{'}$ such that $|m|-|g(x)|<|g(x)-m|<\frac{|m|}{2} \implies |g(x)|>\frac{|m|}{2}$
$|\frac1{g(x)}-\frac1m|=\frac{|g(x)-m|}{|g(x)|\cdot|m|}<\frac{2|g(x)-m|}{m^2}$
There exist a $\delta^{''}$ for $\frac{m^2\cdot\varepsilon}2$ such that $|g(x)-m|<\frac{m^2\cdot\varepsilon}2$
Now for every $\epsilon$ there exists a $\delta=\min\{\delta^{'},\delta{''}\}$
