limit problem

Carsinian Thau

Long story but I need to evaluate:

lim (e^(2x) - 1) / x
x ->0


without using L' Hopital's rule.

I'll be really grateful for any help.

The answer's 2 if that helps.

Carsinian Thau

Just in case it's not clear you can find it at http://i174.photobucket.com/albums/w87/credalysis/Limit.jpg

seandoiler

guess you could write Exp[2x]-1 as a taylor series expansion, divide by x and then evaluate the limit....

Exp[2x]-1 -1 = 2x + 2x^2 +4/3 x^3 +2/3 x^4 + higher order of x

divide by x = 2+2x+....

no take limit as x -> 0 and you get 2

Carsinian Thau

Thanks, you really saved my pride.
I ultimately went with a MacLaurin expansion of the top line because I've never dealt with Taylor's Theorem.
Would this have been obvious to a qualified mathematician?

seandoiler

Carsinian Thau wrote:

Thanks, you really saved my pride.
I ultimately went with a MacLaurin expansion of the top line because I've never dealt with Taylor's Theorem.
Would this have been obvious to a qualified mathematician?

well in this case the maclaurin and taylor series are the same....force of habit to say taylor expand.....of course l'hospital's rule is the clearest way to do it, anybody with a maths background should be able to solve it by simply considering the series expansion...