Limit using L'hospital and maybe another trick
Hi, I've been sitting on this more than an hour and still can't find the solution :(
$\displaystyle \displaystyle\lim_{x \to{\infty}}{\displaystyle{x(}(1+\frac{1}{x})^{x} - {e})}$
I put this in a function grapher and it looks like the result should be $\displaystyle {-}\frac{e}{2}$, but I don't understand why this is true or how do I get to this solution. I assume that L'hospital should be used and supposedly there's another trick?
Thanks in advance
Simplifying the derivation
Actually, you can greatly simplify the derivation if you put y=1/x in the limit.
so as $\displaystyle x\to \inf , \\ y\to 0 $
the equation becomes
$\displaystyle
\lim_{y\to0} \frac{(1+y)^{1/y}-e}{y} $
applying l'hospital's rule, we get
$\displaystyle \lim_{y\to0} (1+y)^{1/y} \frac{y-(1+y)ln(1+y)}{y^2} $ which is
$\displaystyle \lim_{y\to0} (1+y)^{1/y} \lim_{y\to0} \frac{y-(1+y)ln(1+y)}{y^2} $
Since the left hand side is e, we apply l'hospital again to the right side
$\displaystyle e \lim_{y\to0} \frac{-ln(1+y))}{2y} $
Applying l'hospital again, we get
$\displaystyle \frac{-e}{2} $