Proof of limits using epsilon-delta rule

Question

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.

Answer 1

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.

Answer 2

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 $\eta=\min(\color{red}{2},3\epsilon).$

Answer 3

$$ |x-3|<\delta \implies \left|\frac{1}{x}-\frac{1}{3}\right|\le \max\left(\frac{1}{3}-\frac{1}{3+\delta},\frac{1}{3-\delta}-\frac{1}{3}\right)=\frac{\delta}{3-\delta} \to 0. $$