Prove that $\lim_{x\to -3} \frac{1}{x}=-\frac{1}{3}$ using epsilon-delta definition.

Question

I just finished a proof for this problem, but I'm not very confident that I have done it correctly. Any feedback, corrections, or suggestions would be very helpful.

By definition $$\lim_{x\to a}f(x)=L$$ means that $\forall\epsilon>0$, $\exists\delta>0$ such that $0\neq|x-a|<\delta$, and x$\neq$a $\implies |f(x)-L|<\epsilon$.

Prove that $$\lim_{x\to -3} \frac{1}{x}=-\frac{1}{3}$$ using the epsilon-delta definition.

Side Work: Want to show $\exists\delta>0$ such that $0\neq|x+3|<\delta \implies |\frac{1}{x}+\frac{1}{3}|<\epsilon$.

$|\frac{1}{x}+\frac{1}{3}|<\epsilon \implies |\frac{3+x}{3x}|<\epsilon \implies |3+x|<|3x|\epsilon$

Choose $\delta<1$ to get $|x+3|<\delta<1$

So, $-1<x+3<1 $

$\implies -4<x<-2$

$\implies -12<3x<-6$

$\implies |3x|>6$

So we now have: $|3+x|<\delta$ and $|3x|>6$.

$\implies |\frac{3+x}{3x}|<\frac{3+x}{6}<\frac{\delta}{6}\le\epsilon$

So, $\frac{\delta}{6}\le\epsilon \implies \delta\le6\epsilon$

Proof: Let $\epsilon>0$ be given.

Choose $\delta=min(1,6\epsilon) \implies \forall x $ with $0\neq|x+3|<\delta \implies x\in(-3-\delta, -3+\delta)$

So, $x\in(-4,-2)$ because $\delta\le 1$

So, $|3x|>6$

Thus, whenever $0\neq|x+3|<\delta$, we have

$|\frac{1}{x}+\frac{1}{3}|= |\frac{3+x}{3x}|<\frac{\delta}{6} \le 6\epsilon \cdot (\frac{1}{6}) =\epsilon$

$\therefore$ The limit holds by definition.

Answer 1

This step of the side work:

$$ -12 < 3x < -6 \implies |3x| < 6$$

is wrong. For instance, $x=-3$ satisfies the first inequality, but not the second.

But then also this step of the proof:

$$ \frac{|x+3|}{|3x|} < \frac{\delta}{6}$$

is wrong if you are assuming that $|3x| < 6$. If you increase the denominator of a fraction, the fraction becomes smaller, not larger.

But I think fixing one mistake fixes the other. In fact, you know

$$ 3x < -6 \implies |3x| > 6 \implies \frac{1}{|3x|} < \frac{1}{6} $$