1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Prove the Limit

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Use the definition of limits to show that
    [tex]lim _{x\rightarrow 2} \frac{1}{x^{2}} = \frac{1}{4}[/tex]

    2. Relevant equations
    [tex]\forall \epsilon > 0, \exists \delta > 0, x \in D, 0 < |x-2| < \delta \Rightarrow |\frac{1}{x^{2}} - \frac{1}{4}| < \epsilon [/tex]

    3. The attempt at a solution

    [tex]|\frac{1}{x^{2}} - \frac{1}{4}| < \epsilon \Rightarrow |\frac{4-x^{2}}{4x^{2}}| < \epsilon \Rightarrow \frac{|4-x^{2}|}{4x^{2}} < \epsilon [/tex]

    [tex]Restrict: 1 < x < 3 \Rightarrow 4<4x^{2}<36 \Rightarrow \frac{1}{36} < \frac{1}{4x^{2}} < \frac{1}{4} \Rightarrow \frac{|4-x^{2}|}{36} < \frac{|4-x^{2}|}{4x^{2}} < \frac{|4-x^{2}|}{4} < \epsilon[/tex]

    And now I'm stuck. I'm trying to find delta based on epsilon. Normally the problems come to a neat little solution, but |4-x2| is not |x-2| and trying to get it to equal |x-2| is where I'm having the trouble.

    Any help is appreciated.
  2. jcsd
  3. Mar 19, 2009 #2


    User Avatar
    Homework Helper
    Gold Member

  4. Mar 19, 2009 #3
    Yeah I got that far. I know that
    [tex]|(x-2)||(x+2)| < 4\epsilon[/tex]

    But that still leaves that (x+2) there (which I can remove the absolute value from since it has to be positive). I can't have delta dependent on x.

    I'm probably missing something blatant and obvious right here.
    Last edited: Mar 19, 2009
  5. Mar 19, 2009 #4


    User Avatar
    Homework Helper
    Gold Member

    Well, you've restricted your domain to between 1 and 3, so..... __?__[itex]\leq|x+2|\leq[/itex]__?__ .....then simply choose epsilon to be greater than the upper bound of [tex]\frac{|x-2||x+2|}{4}[/tex]
  6. Mar 19, 2009 #5
    ah, duh, thanks. 3 < |x+2| < 5 (since x+2 is positive with or without the absolute value)

    So then

    [tex]\frac{|x-2||x+2|}{4} \leq \frac{5*|x-2|}{4} < \epsilon [/tex]

    Thus [tex]|x-2| < \frac{4}{5}\epsilon[/tex]

    Thus if [tex]\delta = \frac{4}{5}\epsilon \Rightarrow |f(x) - \frac{1}{4}| < \epsilon[/tex]

    Thanks so much.
  7. Mar 20, 2009 #6


    User Avatar
    Homework Helper

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook