how to prove that if given $\lim_{x\to a}f(x)=L$ then $\lim_{x\to a}\frac{1}{f(x)}=\frac{1}{L}$ how to prove that if given $\lim_{x\to a}f(x)=L$ then $\lim_{x\to a}\frac{1}{f(x)}=\frac{1}{L}$
all i have so far is 
$|\frac{f(x)-L}{Lf(x)}|<\epsilon$
And i don't know how to proceed from there
Link to other similar question
I know that the above link has the proof but I was confused why the Lf(x) was on the numerator not on the denominator at the end of the proof and was very confused
 A: The assumption that $L\ne0$ is fundamental here, so I'll proceed under the condition $L\ne0$.
Here's the key: we need a lower bound on $|f(x)|$, which gives an upper bound for $1/|f(x)|$.
The inequality you have to solve is
$$
\left|\frac{1}{f(x)}-\frac{1}{L}\right|<\varepsilon
$$
Since $L\ne0$, there exists $\delta_0$ such that, for $0<|x-a|<\delta_0$,
$$
|f(x)-L|<|L|/2
$$
which easily implies
$$
|f(x)|>\frac{|L|}{2}
$$
and so
$$
\frac{1}{|f(x)|}<\frac{2}{|L|}
$$
With the upper bond at our disposal, we can go on. Take $\delta>0$ such that $\delta\le\delta_0$ and, for every $x$ with $0<|x-a|<\delta$,
$$
|f(x)-L|<\frac{L^2\varepsilon}{2}
$$
Then
$$
\left|\frac{1}{f(x)}-\frac{1}{L}\right|=
\left|\frac{L-f(x)}{Lf(a+h)}\right|=
\frac{1}{L}|f(x)-L|\frac{1}{|f(x)|}<
\frac{1}{L}\frac{L^2\varepsilon}{2}\frac{2}{L}
=\varepsilon
$$

But how did we get to $L^2\varepsilon/2$? It can be determined a posteriori. 
Let me rewrite:
Now take $\delta>0$ such that $\delta\le\delta_0$ and, for every $x$ with $0<|x-a|<\delta$,
$$
|f(x)-L|<E
$$
where $E$ will be determined later.
Then
$$
\left|\frac{1}{f(x)}-\frac{1}{L}\right|=
\left|\frac{L-f(x)}{Lf(a+h)}\right|=
\frac{1}{L}|f(x)-L|\frac{1}{|f(x)|}<
\frac{1}{L}E\frac{1}{L/2}=\frac{2E}{L^2}
$$
and the last term equals $\varepsilon$ as soon as
$$
E=\frac{L^2\varepsilon}{2}
$$
A: We know that $\lim_{x\to a} C=C$, where $C$ is constant and moreover $\lim_{x \to a} \frac{f(x)}{g(x)}= \frac{\lim_{x\to a}f(x)}{\lim_{x \to a} g(x)}$. By using these two facts you can have your result immediately.
A: If $L\neq 0$, then given $\epsilon<|L|$ there is $\delta$ s.t.
$|x-a|<\delta$ implies $|f-L|<\epsilon$. Hence
$$\frac{-\epsilon }{L(L-\epsilon )}\leq
\bigg|\frac{1}{f}- \frac{1}{L}\bigg| = \bigg|\frac{f-L}{fL}\bigg| \leq
\frac{\epsilon }{L(L-\epsilon )}
$$
