In Problems in Mathematical Analysis, Volume 2 by Kaczor and Nowak, Problem 2.3.34 is stated as follow.
Find this limit using L`Hôpital rule
$$\lim\limits_{x\to\infty}x((1+\frac{1}{x})^x-e).$$
I have tried by using L'Hospital rule five times but have no result. I can compute this limit by using Taylor expansion.
First, change the limit into this form,
$$\lim\limits_{x\to 0}\frac{(1+x)^{1/x}-e}{x}.$$
and apply L`Hôpital rule to get,
$$\lim\limits_{x\to 0} \;(x+1)^{1/x} \frac{\left(\frac{x}{x+1}-\log (x+1)\right)}{x^2}.$$
Since $\lim\limits_{x\to 0} \;(x+1)^{1/x} =e$ , you may want to know if $\lim\limits_{x\to 0}\frac{\left(\frac{x}{x+1}-\log (x+1)\right)}{x^2}$ exist. Apply L`Hôpital rule again, $$\lim\limits_{x\to 0}\frac{\left(\frac{x}{x+1}-\log (x+1)\right)}{x^2} = \lim\limits_{x\to 0} \frac{-1}{2 (x+1)^2} = -\frac{1}{2}$$
It exist and we are done,
$$\lim\limits_{x\to 0}\frac{(1+x)^{1/x}-e}{x}=\lim\limits_{x\to 0} \;(x+1)^{1/x} \lim\limits_{x\to 0}\frac{\left(\frac{x}{x+1}-\log (x+1)\right)}{x^2} = -\frac{e}{2}.$$
You can use L'Hopital Rule and substitution to get the answer. In fact \begin{eqnarray} \lim_{x\to\infty}x\left[(1+\frac 1x)^x-e\right]&=&\lim_{x\to\infty}\frac{(1+\frac 1x)^x-e}{\frac 1x} \quad (\text{let } u=\frac 1x)\\ &=&\lim_{u\to 0}\frac{(1+u)^{\frac{1}{u}}-e}{u}\\ &=&\lim_{u\to 0}\frac{e^{\frac{1}{u}\ln(1+u)}-e}{u}\\ &=&\lim_{u\to 0}e(\frac{1}{u}\ln(1+u))'\\ &=&-\frac{e}{2}. \end{eqnarray} Here we use $\ln(1+x)=x-\frac{x^2}{2}+o(x^2)$ to compute $$ \lim_{u\to 0}(\frac{1}{u}\ln(1+u))'=-\frac 12. $$
