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Lim (x-> infinity) sinh(x)sinh(e^(-x))

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


lim (x-> infinity) sinh(x)sinh(e^(-x))


Homework Equations


None really.


The Attempt at a Solution


L'Hospital?
 

Answers and Replies

  • #2
Dick
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l'Hopital, yes. Try writing it as sinh(e^(-x))/(1/sinh(x)). Now the form is 0/0.
 
  • #3
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hmm.. still failing at this question. pls help further?
I don't get a nice solution on applying lhospital
 
  • #4
Dick
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What did you get from l'Hopital?
 
  • #5
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i now get the limit to infinity of:
e^(-x)cosh(e^(-x))/(cothxcosechx)
considering substitution of y=f(x) then computing the limit for lny
 
  • #6
Dick
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That looks ok. Now look at the parts. What's lim coth(x)? What's lim cosh(e^(-x))?
 
  • #7
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cothx -> 1
cosechx -> 0
cosh(e^-x) -> 1
e^-x -> 0
i guess lhopital again...
 
  • #8
Dick
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Not so fast. Aside from the stuff that goes to 1, you've got e^(-x)*sinh(x). What's that? Use the definition of sinh.
 
  • #9
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sweet. so it's 1/2.
 
  • #10
Dick
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Not so bad, huh?
 
  • #11
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Nope. Must be getting tired... it's like midnight here.
 
  • #12
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Also could u please verify that limit (x,y) -> (0,0) of (x^4*y^2)/(x^2 + y^2)2 is 0
 
  • #13
Dick
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Express the function in polar coordinates. Count powers of r.
 

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