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Limit Proof

  • Thread starter ciubba
  • Start date
  • #1
65
2

Homework Statement


Prove [tex]lim_{x->3}\frac{1}{x}=\frac{1}{3}[/tex]

Homework Equations


Epsilon/delta definition

The Attempt at a Solution


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

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

[tex]-\delta+3<x<\delta+3[/tex]

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

[tex]2<x<4 [/tex]

[tex]|x|<1[/tex]

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?
 

Answers and Replies

  • #2
O_o
32
3
Fix
[tex]\epsilon > 0 [/tex] and [tex] \delta = \min\{1, 6\epsilon\}[/tex]
Assume [tex]0 < |x- 3| < \delta [/tex]
then [tex] 0 <|x - 3| < \delta \leq 1 \implies 2 < x < 4 \implies \frac{1}{x} < \frac{1}{2}[/tex]
Now [tex] \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 [/tex]
 
  • #3
65
2
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
 
  • #4
O_o
32
3
If [tex] 2 < x < 4[/tex] then that's the same as saying [tex] 2 < x [/tex] and [tex]x < 4[/tex] You can manipulate inequalities the same way you do equalities so this means [tex]\frac{1}{x} < \frac{1}{2}[/tex] and [tex]\frac{1}{4} < \frac{1}{x}[/tex] 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 [tex]\frac{1}{3} \frac{1}{|x|}[/tex] so if 1/x < 1/2 then [tex]\frac{1}{3}\frac{1}{x} < \frac{1}{3} \frac{1}{2}[/tex]
 
  • #5
65
2
Ah, I didn't know that you could do that with inequalities.

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

Edit: NVM, found it. Thanks!
 

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