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