1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit Proof

  1. Jan 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove [tex]lim_{x->3}\frac{1}{x}=\frac{1}{3}[/tex]

    2. Relevant equations
    Epsilon/delta definition

    3. 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?
     
  2. jcsd
  3. Jan 15, 2015 #2

    O_o

    User Avatar

    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]
     
  4. Jan 15, 2015 #3
    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
     
  5. Jan 15, 2015 #4

    O_o

    User Avatar

    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]
     
  6. Jan 15, 2015 #5
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Limit Proof
  1. Limit proof (Replies: 3)

  2. A limit proof (Replies: 5)

  3. Limit proofs (Replies: 1)

  4. Limit Proof (Replies: 2)

  5. Limit proof (Replies: 4)

Loading...