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Computing limits - logs vs powers

  • Thread starter wombat4000
  • Start date
  • #1
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Homework Statement



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

Homework Equations



powers beat logs

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?
 

Answers and Replies

  • #2
31
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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.
 
  • #3
36
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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:
  • #4
31
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Fine so far. How would you compare x^2 and e^2x? (at least for large x).
 
  • #5
36
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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.
 
  • #6
31
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Sorry if I'm being too vague.

x^2<e^2x so x^2+e^2x<2e^2x which gives you another useful bound :)
 
  • #7
36
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brilliant! - thanks

http://img223.imageshack.us/img223/2827/167uqnbqe2.gif [Broken]
 
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