Proving $\frac{1}{x} \to \frac{1}{3}$ as $x \to 3$

[ciubba]

Homework Statement

Prove $$lim_{x->3}\frac{1}{x}=\frac{1}{3}$$

Homework Equations

Epsilon/delta definition

The Attempt at a Solution

$$|\frac{1}{x}-\frac{1}{3}|<\epsilon \; \; \mbox{when} \; \; |x-3|<\delta$$
I expanded the left to get
$$-\epsilon+\frac{1}{3}<\frac{1}{x}<\epsilon+\frac{1}{3}$$

I can't turn that into something of the form x-3 without introducing new solutions, so I tried to expand the right side

$$-\delta+3<x<\delta+3$$

Which didn't help, so I tried defining ϵ<2/3 so that |x-3|<1, so

$$2<x<4$$

$$|x|<1$$

Unfortunately, I don't see any way to turn x into 1/x without inverting the inequality, at which point I'd have > symbols, which doesn't agree with the left side. Any suggestions?

 

[O_o]

Fix
$$\epsilon > 0$$ and $$\delta = \min\{1, 6\epsilon\}$$
Assume $$0 < |x- 3| < \delta$$
then $$0 <|x - 3| < \delta \leq 1 \implies 2 < x < 4 \implies \frac{1}{x} < \frac{1}{2}$$
Now $$\left| \frac{1}{x} - \frac{1}{3}\right| = \frac{1}{|3x|} | 3 - x | = \frac{1}{|3x|} | x - 3| < \frac{1}{3}\frac{1}{2} \delta \leq \frac{1}{6} 6\epsilon = \epsilon$$

 

[ciubba]

I don't understand how you transitioned from 2<x<4 to 1/x<1/2

I also don't see from where you get the 1/3 in <1/3*1/2

 

[O_o]

If $$2 < x < 4$$ then that's the same as saying $$2 < x$$ and $$x < 4$$ You can manipulate inequalities the same way you do equalities so this means $$\frac{1}{x} < \frac{1}{2}$$ and $$\frac{1}{4} < \frac{1}{x}$$ We only need the 1/2 term since we're looking for an upper bound on 1/x. The 1/3 comes from the fact that we have $$\frac{1}{3} \frac{1}{|x|}$$ so if 1/x < 1/2 then $$\frac{1}{3}\frac{1}{x} < \frac{1}{3} \frac{1}{2}$$

 

[ciubba]

Ah, I didn't know that you could do that with inequalities.

I still cannot find $$\frac{1}{3} \frac{1}{|x|}$$ anywhere in the proof. Can you point out on which step that relationship can be found?

Edit: NVM, found it. Thanks!