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**walleye** This is actually a problem from a subject thats mainly involved with Laplace transforms.

it starts off with expanding f(s) = $\displaystyle ln(1+1/s)$ around infinity (by taylor series)

so firstly, evaluating $\displaystyle ln(1+1/s)$ at s = infinity, which tends towards log (1 + 0 ) = 0

then the next term of the taylor series involves f ' (s) evaluated at infinity

$\displaystyle f'(s) = -1 / (s^2 + s )$

$\displaystyle f ' (inf) = -1 / (inf) = 0$ .... right?

f '' (s) = $\displaystyle (2s + 1) / (s^2 + s)^2$

now for evaluating this one at infinity directly would yield inf / inf = 1 ?

but i know (of) l'Hopital's rule

which i tried

so lim s -> inf of $\displaystyle (2s + 1) / (s^2 + s)^2$

= lim s -> inf of $\displaystyle 2 / 4s(s^2 + s) = 0$

here i derived both top and bottom of the fraction (seperately) with respect to s

long story short, if you use l'hopitals rule for the following terms as well, you get 0 for every term in the taylor series, which im assuming isnt right.

am i using l'hopitals correctly? any other help?

also any tips on using the Math-notation available here? :P