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 question

  1. Nov 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Use the precise definition to show
    [tex]lim (x^2+3x) = 10[/tex]

    3. The attempt at a solution

    Let [tex]\epsilon[/tex] > 0

    [tex]x^2 + 3x - 10 < \epsilon [/tex]

    [tex](x-2)^2 = x^2 - 4x + 4 [/tex]

    This doesn't equal the equation. Add 7x, -14

    [tex]\left| x-2 \right| ^2 + 7 \left| x-2 \right| [/tex]

    So far it's alright. Now I need to get a value for [tex]\delta[/tex]

    [tex]\epsilon[/tex] = [tex]\delta^2 + 7 \delta[/tex]

    Now I'm totally confused. Normally I've used simply [tex]\delta[/tex] expressions like [tex]\delta[/tex] = [tex]\epsilon[/tex]/2. What should I say my [tex]\delta[/tex] is equal to in this case, and why?

    So [tex]\left| (x^2 + 3x) -10 \right|[/tex] < [tex]\delta^2 + 7 \delta[/tex]

    Any help?
  2. jcsd
  3. Nov 7, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    It would be better to use the fact that x2+ 3x- 10= (x- 2)(x+ 5) so
    |x2+ 3x- 10|= |x-2||x+ 5| and you want that less than [itex]\epsilon[/itex]. That will be true if [itex]|x-2|< \epsilon/|x+5|[/itex] but you want a constant. If [itex]\gamma> |x+5|[/itex] then [itex]\epsilon/\gamma< \epsilon/|x+5|[/itex] so you can set [itex]\delta= \epsilon/\gamma[/itex]. To find an upper bound on |x+ 5| remember that you are taking x close to 2 anyway: say |x- 2|< 1 so -1< x-2< 1. How large can x+ 5 be?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Limit proof question
  1. Limit question; Proof. (Replies: 10)