# How to evaluate $\lim\limits_{x \rightarrow +\infty}{e^x (e - (1+\frac{1}{x} )^x)}$ without L'Hospital?

Using several times L'Hospital Rule I got $$\lim_{x \rightarrow +\infty}{e^x \left (e - \left(1+\dfrac{1}{x}\right )^x\right)} = +\infty.$$ Is it possible find this limit without L'Hospital?

-
It is always possible to find a limit without l'Hopital as at some point l'Hopital was proving (hopefully without invoking l'Hopital). – Fabian Aug 15 '12 at 20:49
what we allowed to use? – Norbert Aug 15 '12 at 20:55
The 0% accept rate does not look inviting... – Argon Aug 15 '12 at 21:07
Yeah, that accept rate would cause even $\,\lim_{n\to\infty}1/n\,$ to equal to -4828939045...terrible thing. – DonAntonio Aug 15 '12 at 23:49
You are right. I never watched that. Thanks. – jon jones Aug 16 '12 at 0:17

The natural thing to do is to look at the logarithm of $\left(1+\frac{1}{x}\right)^x$, that is, at $x\log\left(1+\frac{1}{x}\right)$. Use the series $$\log(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots.$$ From this we can obtain good estimates of the difference between $e$ and $(1+1/x)^x$ when $x$ is large. For the calculation, the series expansion of $e^t$ is useful.