TITLE: Finding a limit involving exponential function
QUESTION [6 upvotes]: Find the value of the limit $$\lim_{x\to0}\left(\frac{e-(1+x)^\frac{1}{x}}{x}\right).$$

I tried to apply the standard limit $$\lim_{x\to0}(1+x)^\frac{1}{x} = e$$ and L'Hospital's Theorem individually, but that didn't help me. 
Any help will be appreciated.

REPLY [3 votes]: Put $$f (x)=(1+x)^\frac 1x $$ for $x\ne 0$ and $f (0)=e $.
then your limit is
$$-\lim_0\frac {f (x)-f (0)}{x-0} $$
or
$$-\lim_0 f'(x ).$$
with
$f'(x)=f (x)(-\frac {1}{x^2}\ln (1+x)+\frac {1}{x (1+x)}) $
$=f (x)(\frac {-x+x^2/2 (1+\epsilon (x))}{x^2}+\frac {1}{x}-\frac {1}{1+x}) $
from this, the limit is $$-e(\frac {1}{2}-1)=\frac {e}{2} $$