Evaluate a limit (probably involving L'Hôpital rule) Evaluate the limit:
$$\mathop {\lim }\limits_{x \to \infty } x\left( {{{\left( {1 + {1 \over x}} \right)}^x} - e} \right)$$
My attempts didn't yield a result. I'd be glad for a guidance.
Thanks!
 A: There is a nice little trick to do this.
$$\lim\limits_{h\rightarrow 0+}{\frac{1}{h}\left((1+h)^{\frac{1}{h}}-e\right)}$$
$$(1+h)^{\frac{1}{h}}=\exp{\left(\frac{1}{h}\ln{(1+h)}\right)}=\exp\left({\frac{h-\frac{h^2}{2}+o(h^2)}{h}}\right)=e\left(1-\frac{h}{2}+o(h)\right)$$
Hence the limit is reduced to.
$$\frac{1}{h}\left((1+h)^{\frac{1}{h}}-e\right)=e\left(-\frac{1}{2}+o(1)\right)\longrightarrow -\frac{e}{2}$$ as $h\rightarrow 0^+$.
A: You just have to persist in the L'Hopital rule. 
Start with
$$\frac{(1+\frac{1}{x})^x-e}{\frac{1}{x}}$$ top and bottom go to zero,
by L'Hopital we get
$$\frac{(\ln(1+\frac{1}{x})-\frac{1}{x+1})(1+\frac{1}{x})^x}{-\frac{1}{x^2}}
=(\frac{x^2}{x+1}-x^2\ln(1+\frac{1}{x}))(1+\frac{1}{x})^x$$
Now $(1+\frac{1}{x})^x\to e$ so lets set that term aside.
What remains can be written
$$\frac{\frac{1}{x+1}-\ln(1+\frac{1}{x})}{\frac{1}{x^2}}$$
top and bottom go to zero so again with L'Hopital,
$$\frac{-\frac{1}{(x+1)^2}+\frac{x}{x+1}\frac{1}{x^2}}{-\frac{2}{x^3}}=-\frac{1}{2}(\frac{x^2}{x+1}-\frac{x^3}{(x+1)^2})=-\frac{1}{2}\frac{x^2}{(x+1)^2}\rightarrow -\frac{1}{2}$$
So we get the limit $-\frac{e}{2}$
A: Ugh @ L'Hôpital's Rule :)
Let $f(h) = \begin{cases} (1+h)^{1/h}, & h\ne 0 \\ e, & h=0 \end{cases}$. Then the limit you want is $f'(0)$.
(To verify this, we note, substituting $h=1/x$, that $$\lim_{x\to\infty} x\left(\Big(1+\frac1x\Big)^x - e\right) = \lim_{h\to 0^+} \frac{(1+h)^{1/h}-e}h = \lim_{h\to 0^+} \frac{f(h)-f(0)}h.)$$
Using a Taylor polynomial, one checks that $\ln f(h) = \dfrac1h\ln(1+1/h) = 1-\frac12 h + \epsilon(h)$, where $\lim\limits_{h\to 0}\dfrac{\epsilon(h)}h=0$, so
$$\frac{f'(0)}{f(0)} = (\ln f)'(0) = -\frac12.$$
Therefore, $f'(0) = -e/2$. 
