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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.

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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

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

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  • $\begingroup$ Ah, of course it was something silly. Take a look at the edit I made. $\endgroup$ – Dallas W. Jan 29 at 3:01
  • $\begingroup$ Yes, now it looks right. Although, editing the question makes my answer out of sync. It would be more aligned with the Stack Exchange guidelines to leave the question as it was, then write your own answer with the correct version. $\endgroup$ – Matthew Leingang Jan 29 at 16:04

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