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.

• Correct. It's a good practice to try different numbers.
– user399481
Commented Dec 24, 2016 at 16:24
• You did the first to next to last line perfectly but you need to add to the last line an introductory clause that you are assume $|x-3| < \delta$ in the first place. e.g. "Whenever $|x - 3| < \delta$ and $x >$ then $|\frac 1x - 13| < \frac {|x-3|}3 < \frac{\delta} 3 < \epsilon$". Ideally you want to do the proof directly "For epsilon let delta = .... When |x-a| < delta then .... |f(x) - c| < epsilon" but as we have to calculate delta we usually do these proofs backwards as you did. Which is fine but then we need to recap a statement that the argument works in the forward direction. Commented Dec 24, 2016 at 16:43
• @juniven. Why? What the OP did is already much more general and correct. Specific numbers are good to get inspiration but the OP has already figured it out completely. Commented Dec 24, 2016 at 16:45
• Epsilon-delta is not a rule.
– user9464
Commented Dec 24, 2016 at 16:46

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

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.

• But $|x-3|<1 \not \implies 1<x<4$ Commented Dec 24, 2016 at 17:06

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