L'Hopital's Rule?

Question

Help with this?

http://www.geocities.com/jsprplc2006/Picture1.png

They say you have to use L'hopital's rule.

Answer 1

I'm sure you mean as x->inf in your picture.

L'Hopital's rule helps with limits where both the top and bottom go to 0 or both go to infinity.

Essentially it says that lim (f/g) = lim (df/dx / dg/dx).

So in your problem, the top and bottom both go to 0...

lim x->inf (1/x / (2^(1/x) - 1)) =
lim x->inf ((-1/x^2) / ((-1/x^2) * ln(2) *2^(1/x)) =
lim x->inf(1/((ln(2) * 2^(1/x))) = 1 / ln(2)

Answer 2

Let y = 1/x. Then we are interested in the limit as y -> 0 of y/(2*y - 1). Since at 0 this would be 0/0, l'Hopital's rule applies, and we differentiate both numerator and denominator to obtain 1/((2*y) ln 2) Both numerator and denominator are finite at y = 0, so we are done, getting 1/(ln 2) as the limit. Which, of course, is the limit of the original expression as well.

Answer 3

The function approaches 0/0, so we can use l'hopital's rule. It states that when a function approaches 0/0 or infinity/infinity, we can take the derivative of the top and the bottom and find the limit. You keep taking the derivative until you get a real limit.

Answer 4

When you have a limit in indeterminant form, such as 0/0 you can use L'Hospital's rule.

lim x → ∞ of (1/x)/{2^(1/x) - 1}
= lim x → ∞ of (-1/x²)/{(-1/x²)(ln 2)2^(1/x)}
= lim x → ∞ of 1/{(ln 2)2^(1/x)}
= 1/{(ln 2)*1} = 1/(ln 2)