Find the limit

• November 16th 2008, 12:49 PM
dm10
Find the limit
Right now I'm doing some problems that requires you to find the limit using the L'hospital's rule. But our teacher told us not to use the rule for this problem:

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

I'm not sure on how to do this though. The answer I get is 0/0 but I know that's not correct.
• November 16th 2008, 01:06 PM
Jhevon
Quote:

Originally Posted by dm10
I'm sorry but I don't understand what you just said.

for example,

$e^x = \sum_{n = 0}^\infty \frac {x^n}{n!} = 1 + x + \frac {x^2}{2!} + \frac {x^3}{3!} + \cdots$ for all $x$

with that, you can write the power series for $e^{-x}$

Moreover, $\sin x = \sum_{n = 0}^\infty (-1)^n \frac {x^{2n + 1}}{(2n + 1)!} = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} +- \cdots$ for all $x$

so you can rewrite each expression as power series. see what cancels out and continue that way.