Limits involving natural exponential2

  • Thread starter synergix
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  • #1
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Homework Statement



lim(e-2x cosx)
x-> infinity

The Attempt at a Solution



I thought right away that the limit would not exist because of cos x oscillating the function between + and - but the answer in the book says zero. I need help figuring out why my thinking was incorrect.
 

Answers and Replies

  • #2
Dick
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-1<=cos(x)<=1. Whether there's a limit depends on what e^(-2x) does. What does it do? For a proof think about using the squeeze theorem.
 
  • #3
lanedance
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think about what happens to the negative exponential as x gets large, and maybe try a squeeze theorem
 
  • #4
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e^(-2x) will get very small.
 
  • #5
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well it will still be oscillating but approaching zero all the same I suppose
 
  • #6
Dick
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e^(-2x) will get very small.

Ok, so what's you conclusion? And why?
 
  • #7
Dick
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well it will still be oscillating but approaching zero all the same I suppose

That's fine, if you don't have to prove it and omit the 'I suppose'.
 
  • #8
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How would I apply the squeeze theorem to this problem?
 
  • #9
Dick
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How would I apply the squeeze theorem to this problem?

Can you find two functions f(x) and g(x) such that f(x)<=cos(x)e^(-2x)<=g(x) such that f(x) and g(x) both approach zero? Possibly using -1<=cos(x)<=1?
 
  • #10
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Because
-1<=cos(x)<=1

-e-2x <= cos(x)e-2x <= e-2x

is that it? I just emulated what I have seen on a few math help sites. But it makes sense now that I have thought it out a bit.

soo

lim -e-2x=0=lim e -2x
x->infinity x->infinity

so

lim cos(x)e-2x=0
x->infinity
 
  • #11
Dick
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Because
-1<=cos(x)<=1

-e-2x <= cos(x)e-2x <= e-2x

is that it? I just emulated what I have seen on a few math help sites. But it makes sense now that I have thought it out a bit.

soo

lim -e-2x=0=lim e -2x
x->infinity x->infinity

so

lim cos(x)e-2x=0
x->infinity

Brilliant. A little emulation goes a long way.
 

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