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Computing limits - logs vs powers

  1. May 18, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\frac{log (x^{2} + e^{2x})}{x + 3}[/tex] find the limit as x_> infinty

    2. Relevant equations

    powers beat logs

    3. The attempt at a solution

    going by the powers beat logs idea - simply, the limit as n-> infinity is 0.

    is this correct? can you simply say that powers beat logs always?
  2. jcsd
  3. May 18, 2008 #2
    No it's not that simple. Intuitively e^2x is going to dominate the bracket in the numerator and log(e^2x) can be easily simplified. Can you sandwich x^2+e^2x between multiples of e^2x (at least for large x)? Try the sandwich theorem.
  4. May 18, 2008 #3
    ok - it will go to 2 then?

    [tex]\frac{log (e^{2x})}{x + 3} \leq \frac{log (x^{2} + e^{2x})}{x + 3}
    \leq ? [/tex]

    [tex]\frac{log (e^{2x})}{x + 3} = \frac{2x}{x + 3} [/tex] which tends to 2 as x tends to infinity

    is this correct - what should i use for thew upper bound?
    Last edited: May 18, 2008
  5. May 18, 2008 #4
    Fine so far. How would you compare x^2 and e^2x? (at least for large x).
  6. May 18, 2008 #5
    e^2x is considerably larger than x^2 ?

    if x^2 divereges - so must e^2x?

    cant say im 100% sure what your ginting towards.
  7. May 18, 2008 #6
    Sorry if I'm being too vague.

    x^2<e^2x so x^2+e^2x<2e^2x which gives you another useful bound :)
  8. May 18, 2008 #7
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