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

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In summary, the conversation discusses using L'Hospital's rule to find the limit of a function as x approaches infinity, using substitution and the definition of sinh. The limit is found to be 1/2. The conversation also briefly touches on finding a limit in polar coordinates.
  • #1
wimma
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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?
 
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  • #2


l'Hopital, yes. Try writing it as sinh(e^(-x))/(1/sinh(x)). Now the form is 0/0.
 
  • #3


hmm.. still failing at this question. pls help further?
I don't get a nice solution on applying lhospital
 
  • #4


What did you get from l'Hopital?
 
  • #5


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


That looks ok. Now look at the parts. What's lim coth(x)? What's lim cosh(e^(-x))?
 
  • #7


cothx -> 1
cosechx -> 0
cosh(e^-x) -> 1
e^-x -> 0
i guess lhopital again...
 
  • #8


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


sweet. so it's 1/2.
 
  • #10


Not so bad, huh?
 
  • #11


Nope. Must be getting tired... it's like midnight here.
 
  • #12


Also could u please verify that limit (x,y) -> (0,0) of (x^4*y^2)/(x^2 + y^2)2 is 0
 
  • #13


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

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

1. What is the limit of sinh(x) as x approaches infinity?

The limit of sinh(x) as x approaches infinity is also infinity. This is because the hyperbolic sine function grows exponentially as x increases.

2. What is the limit of sinh(e^(-x)) as x approaches infinity?

The limit of sinh(e^(-x)) as x approaches infinity is 0. This is because as x approaches infinity, e^(-x) approaches 0 and sinh(0) = 0.

3. What is the product of two limits?

The product of two limits is equal to the limit of the product. In this case, the limit of sinh(x) as x approaches infinity multiplied by the limit of sinh(e^(-x)) as x approaches infinity is equal to infinity multiplied by 0, which is equal to 0.

4. How does the limit of sinh(x)sinh(e^(-x)) as x approaches infinity compare to the limit of sinh(x) as x approaches infinity?

The limit of sinh(x)sinh(e^(-x)) as x approaches infinity is equal to 0, while the limit of sinh(x) as x approaches infinity is equal to infinity. This means that the limit of the product is significantly smaller than the limit of sinh(x) alone.

5. What does this limit tell us about the behavior of the function?

This limit tells us that as x approaches infinity, the function sinh(x)sinh(e^(-x)) approaches 0. This means that the value of the function gets closer and closer to 0 as x gets larger, but never actually reaches 0. This behavior is known as asymptotic behavior.

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