Proof of limits using epsilon-delta rule Prove: $\lim_{x\to 3}\frac 1x=\frac 13$;
According to the epsilon-delta rule if
$|f(x)-L|<\epsilon$ whenever $|x-a|<\delta$, then $\lim_{x\to a}f(x)=L$.
Steps I Followed:
$\left|\frac1x-\frac13\right|=\frac{|3-x|}{3|x|}=\frac{|x-3|}{3|x|}<\frac{|x-3|}{3}$ as $x\to3$, i.e. $x>1$
Now, if we choose $\delta=3\epsilon$
we can write $\left|\frac1x-\frac13\right|<\frac{|x-3|}{3}<\frac{\delta}{3}=\epsilon\\$
which ends our proof.
I would like to know if the steps performed in the first line are correct, i.e. whether we can use this sort of assumption here. 
 A: Try this. Maybe this can help.
Let $\epsilon >0$. If $|x-3|<1$, then we get
$$2<x<4$$
and so with this, we can take $\delta=\min\{6\epsilon,1\}$. Thus, if $|x-3|<\delta$, then
\begin{align}
\left|\frac{1}{x}-\frac{1}{3}\right|&=\frac{|x-3|}{3|x|}\\
&=\frac{|x-3|}{3x}\\
&<\frac{\delta}{6}\\
&\leq \epsilon.
\end{align}
The result follows.
A: Given an $\epsilon>0$, we look for $\eta>0 $ such that
$$|x-3|<\eta \implies |\frac{1}{x}-\frac{1}{3}|<\epsilon$$
as $x$ tends to $3$, we can suppose that $2<x<4$. then $ |x-3|<\color{red}{1} $ and $\frac{1}{4}<\frac{1}{x}<\frac{1}{2}$.
thus
$$|\frac{1}{x}-\frac{1}{3}|=\frac{|x-3|}{3x}<\frac{1}{6}|x-3|$$
Now, we see that we can take $\eta=\min(\color{red}{1},6\epsilon)$.
Instead of the condition $2<x<4$, one could take $1<x<5$. In this case he will have $|x-3|<\color{red}{2}$ and
$\frac{1}{3x}<\frac{1}{3}$.
So he will take $\eta=\min(\color{red}{2},3\epsilon).$
A: $$
|x-3|<\delta \implies \left|\frac{1}{x}-\frac{1}{3}\right|\le \max\left(\frac{1}{3}-\frac{1}{3+\delta},\frac{1}{3-\delta}-\frac{1}{3}\right)=\frac{\delta}{3-\delta} \to 0.
$$
