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?
 A: 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. 
A: We need to prove that,
$$\lim_{x \rightarrow +\infty}{e^x \left (e - \left(1+\dfrac{1}{x}\right )^x\right)} = +\infty.$$
consider
$$\lim_{x \rightarrow +\infty}{e^x \left (e - M\right)} $$
where, $$M = \left(1+\dfrac{1}{x}\right )^x$$
if we prove that $M$ has a finite limit, we are done.
Note that, 
1. M is increasing function of x
2. M is bounded above

first one you can prove as an exercise, for second
$$M = \left(1+\dfrac{1}{x}\right )^x = \left(\left(1+\dfrac{1}{x}\right )^{x/k} \right)^k < \left(\frac{1}{1-\frac{1}{x} \cdot\frac{x}{k}}\right)^k = \left ( \frac{1}{\left(1-\frac{1}{k}\right)^k}\right)$$
so that, $$M< \frac{1}{\left(1-\frac{1}{k}\right)^k}$$ for any whole k.
$$\lim_{x \rightarrow +\infty}{e^x \left (e - M\right)} =  \lim_{x \rightarrow +\infty}{e^x L} = +\infty $$
