Finding the limit of $s((1+\frac{1}{s})^{s} - e)$ as $s$ aproaches infinity I came across that limit:
$\lim_{s\to\infty} s\bigg(\big(1+\frac{1}{s}\big)^{s} - e\bigg)$
I tried to solve it using l'Hospital's rule:
$\lim_{s\to\infty} s\bigg(\big(1+\frac{1}{s}\big)^{s} - e\bigg) = \lim_{t\to0} \frac{1}{t} \bigg(\big(1+t\big)^{\frac{1}{t}} - e\bigg) = \lim_{t\to0} e^{\frac{1}{t}\log(1+t)}\big(\frac{1}{t(t+1)} - \frac{\log(1+t)}{t^{2}}\big)$
I used l'Hospital's rule after the second equality.
However I got nothing really valuable. Do you have any ideas how to solve that problem?
 A: Note that we can write
$$\begin{align}
\left(1+\frac1s\right)^s&=e^{s\log\left(1+\frac1s\right)}\\\\
&=e^{\left(1-\frac{1}{2s}+O\left(\frac1{s^2}\right)\right)}\\\\
&=e\left(1-\frac{1}{2s}+O\left(\frac1{s^2}\right)\right)
\end{align}$$
Therefore, 
$$\begin{align}
\lim_{s\to \infty}\left(s\left(\left(1+\frac1s\right)^s-e\right)\right)&=\lim_{s\to \infty}\left(-\frac12 e+O\left(\frac1s\right)\right)\\\\
&= -\frac12 e
\end{align}$$
A: If you dont want to write all those ugly oh's all over the place you could write your expression as
$$es\frac{e^{s\ln(1+\frac{1}{s})-1}-1}{ s\ln(1+\frac{1}{s})-1}
(s\ln(1+\frac{1}{s})-1)$$
and then
$$s(s\ln(1+\frac{1}{s})-1)\to -\frac{1}{2}$$
is the well known
$$\lim\limits_{x\to 0}\frac{\ln (1+x)-x}{x^2}=-\frac{1}{2}$$
A: Let us consider $$A= \frac{1}{t} \bigg(\big(1+t\big)^{\frac{1}{t}} - e\bigg) $$ where $t$ is small compared to $1$.
First, let us look at $$B=\big(1+t\big)^{\frac{1}{t}}\implies \log(B)=\frac{1}{t}\, \log(1+t)$$ Now, using Taylor expansions $$\log(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}+O\left(t^4\right)$$ $$\log(B)=1-\frac{t}{2}+\frac{t^2}{3}+O\left(t^3\right)$$ $$B=e^{\log(B)}=e-\frac{e t}{2}+\frac{11 e t^2}{24}+O\left(t^3\right)$$
I am sure that you can take it from here and find not only the limit but also how it is approached.
A: To sum up the topic and post the full solution:
$\lim\limits_{x\to \infty}s\bigg(\big(1+\frac{1}{s}\big)^{s}-e\bigg) = \lim\limits_{x\to \infty}\frac{\big(1+\frac{1}{s}\big)^{s}-e}{\frac{1}{s}} = \lim\limits_{x\to \infty}\frac{\big(1+\frac{1}{s}\big)^{s}\bigg(\log(1+\frac{1}{s})-s\frac{\frac{1}{s^{2}}}{1+\frac{1}{s}}\bigg)}{\frac{-1}{s^{2}}} = \lim\limits_{x\to \infty}\frac{\big(1+\frac{1}{s}\big)^{s}\big(\frac{1}{s+1}-\log(1+\frac{1}{s})\big)}{\frac{1}{s^{2}}} = \lim\limits_{x\to \infty}\big(1+\frac{1}{s}\big)^{s}\cdot\lim\limits_{x\to \infty}\bigg(\frac{\frac{1}{s+1}-\log(1+\frac{1}{s})}{\frac{1}{s^{2}}}\bigg)$
Now let's calculate only the second limit using l'Hospital's rule:
$\lim\limits_{x\to \infty}\bigg(\frac{\frac{1}{s+1}-\log(1+\frac{1}{s})}{\frac{1}{s^{2}}}\bigg)= \lim\limits_{x\to \infty}\frac{\frac{1}{(s+1)^{2}}-\frac{1}{s(s+1)}}{\frac{2}{s^{3}}}=\frac{-1}{2}\lim\limits_{x\to \infty}\frac{s^{2}}{s(s+1)^{2}}=\frac{-1}{2}$
Thus:
$\lim\limits_{x\to \infty}s\bigg(\big(1+\frac{1}{s}\big)^{s}-e\bigg)=\frac{-e}{2}$
That's a pretty nice solution I think.
