$\epsilon$-$\delta$ proof, $\lim\limits_{x \to a}$ $\frac{1}{x}$ = $\frac{1}{a}.$ It's been a while since I've studied $\epsilon$-$\delta$ proofs so I'm trying to get a good understanding of how to go about solving this.
Working backwards using this as a reference: $\epsilon$-$\delta$ proof that $\lim\limits_{x \to 1} \frac{1}{x} = 1$.
Solve:

$\lim\limits_{x \to a}$ $\frac{1}{x}$ = $\frac{1}{a}.$

We assume:

$\lim_{x\to a}f(x)=\ell \neq 0$.

And have:

$\left|\frac{1}{f(x)}-\frac 1{\ell}\right|=\frac{|f(x)-\ell|}{|f(x)||\ell|}$

Now suppose that a = 4.
So:

$\left|\frac{1}{(\frac{1}{x})}-\frac 1{(\frac{1}{4})}\right|=\frac{|\frac{1}{x}-\frac{1}{4}|}{|\frac{1}{x}||\frac{1}{4}|}$

Which yields:

$\left|x-4\right|={|\frac{1}{x}-\frac{1}{4}|}*|{x|*|4|}$

I am confused as to how to go about solving for epsilon or delta at this point.
 A: Given $\epsilon>0$, we need to find $\delta$ so that $\left|\frac 1x -\frac 1 a\right|<\epsilon$ for $|x-a|<\delta$.
Now $$\left|\frac 1x -\frac 1 a\right|=\frac{|x-a|}{|xa|}.$$
What we are going to do is pick a $\delta$ that constrains both $|x-a|$ and $\frac{1}{|xa|}$.
First, we want to control the latter. If $|x-a|<\frac{|a|}{2}$ then $|x|>\frac{|a|}{2}$, and thus $|xa|>\frac{|a^2|}{2}$.
So, in this circumstance, you have:
$$\left|\frac 1x -\frac 1 a\right|\leq\frac{2|x-a|}{a^2}.$$
So we set $$\delta =\min\left(\frac{|a|}{2},\frac{\epsilon a^2}{2}\right)$$
Now if $|x-a|<\delta,$ then $|x-a|<\frac{\epsilon a^2}{2}$ and $|xa|>\frac{a^2}{2}$, so:
$$\left|\frac 1x -\frac 1 a\right|<\epsilon$$

This is a fairly common trick in proofs for functions $f(x)$ that go to infinity close to some $x_0$. Given an $a\neq x_0$, you first need to pick a $\delta_0$ so that the interval $[a-\delta_0,a+\delta_0]$ does not include $x_0$. Then you pick a $\delta_1$ so that it works with $\delta_0$ to make $|f(x)-f(a)|<\epsilon$.
A: Let $\epsilon>0$ be given. We have to find a number $\delta>0$ such that $\left|\frac{1}{x}-\frac{1}{a}\right|<\epsilon$ whenever $|x-a|<\delta$. But
$$
\left|\frac{1}{x}-\frac{1}{a}\right|=\left|\frac{a-x}{xa}\right|=\frac{|x-a|}{|xa|}.
$$
We find a positive constant $C$ such that $\frac{1}{|xa|}<C\Rightarrow\frac{|x-a|}{|xa|}<C|x-a|$, and we can make $C|x-a|<\epsilon$ by taking $|x-a|<\frac{\epsilon}{C}=\delta$. We restrict $x$ to lie in the interval $|x-a|<1$ which implies that $a-1<x<a+1$ so that
$$
\frac{1}{a-1}>\frac{1}{x}>\frac{1}{a+1}\Rightarrow\frac{1}{a(a-1)}<\frac{1}{xa}<\frac{1}{a(a+1)}\Rightarrow\frac{1}{|xa|}<\frac{1}{a(a+1)}.
$$
So $C=\frac{1}{a(a+1)}$ is suitable. Thus, we should choose $\delta=\min\left\{1,\epsilon\cdot a(a+1)\right\}$.
To verify this choice of $\delta$, consider the following:
Given $\epsilon>0$, we let $\delta=\min\left\{1,\epsilon\cdot a(a+1)\right\}$. If $|x-a|<\delta$, then $|x-a|<1\Rightarrow\frac{1}{|xa|}<\frac{1}{a(a+1)}$ and also $|x-a|<\epsilon\cdot a(a+1)$. Hence,
$$
\left|\frac{1}{x}-\frac{1}{a}\right|=\frac{|x-a|}{|xa|}<\frac{1}{a(a+1)}\cdot[\epsilon\cdot a(a+1)]=\epsilon,
$$
as desired. $\blacksquare$
A: $\left|\dfrac{1}{x} - \dfrac{1}{4}\right|= \dfrac{|x-4|}{4|x|}$. In addition, you want $|x-4| < 1\implies -1 < x-4 < 1\implies x > 3 \implies |x| = x > 3\implies \left|\dfrac{1}{x} - \dfrac{1}{4}\right| < \dfrac{1}{12}|x-4|< \epsilon \iff |x-4| < 12\epsilon\implies \delta = \text{min}(1,12\epsilon)$ will do the job !
