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

[wimma]

Homework Statement

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

Homework Equations

None really.

The Attempt at a Solution

L'Hospital?

 

[Dick]

l'Hopital, yes. Try writing it as sinh(e^(-x))/(1/sinh(x)). Now the form is 0/0.

 

[wimma]

hmm.. still failing at this question. pls help further?
I don't get a nice solution on applying lhospital

 

[Dick]

What did you get from l'Hopital?

 

[wimma]

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

 

[Dick]

That looks ok. Now look at the parts. What's lim coth(x)? What's lim cosh(e^(-x))?

 

[wimma]

cothx -> 1
cosechx -> 0
cosh(e^-x) -> 1
e^-x -> 0
i guess lhopital again...

 

[Dick]

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.

 

[wimma]

sweet. so it's 1/2.

 

[Dick]

Not so bad, huh?

 

[wimma]

Nope. Must be getting tired... it's like midnight here.

 

[wimma]

Also could u please verify that limit (x,y) -> (0,0) of (x^4*y^2)/(x^2 + y^2)2 is 0

 

[Dick]

Express the function in polar coordinates. Count powers of r.