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

• wimma

## Homework Statement

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

None really.

## The Attempt at a Solution

L'Hospital?

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

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

What did you get from l'Hopital?

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

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

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

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.

sweet. so it's 1/2.

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

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

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