How to find $ \lim_{x\rightarrow 0} \frac{(x + 1)^{\frac{1}{x}} - e}{x} $ Let
$$ \lim_{x\rightarrow 0} \frac{(x + 1)^{\frac{1}{x}} - e}{x} = ? $$
How would you calculate it's limit? I thought using l'hopital rule, but it then becomes something nasty, as the differentiate of $(x + 1)^{\frac{1}{x}}$ isn't simple.
 A: Hint: $(1+x)^{1/x} = e^{(\ln(1+x))/x}.$ Use $\ln (1+h) = h-h^2/2 +O(h^3), e^h = 1 +h+O(h^2)$ for small $h.$
A: As @zhw already described, an efficient way to approach this is to use asymptotic analysis.  So, here we will use instead the "brute force" approach via L'Hospital's Rule.  To that end, we have
$$\begin{align}
\lim_{x\to 0}\frac{(x+1)^{1/x}-e}{x}&=\lim_{x\to 0}(x+1)^{1/x}\left(\frac{x-(x+1)\log(x+1)}{x^2(x+1)}\right)\\\\
&=e\lim_{x\to 0}\left(\frac{x-(x+1)\log(x+1)}{x^2(x+1)}\right)\\\\
&=e\lim_{x\to 0}\left(\frac{-\log(x+1)}{3x^2+2x}\right)\\\\
&=e\lim_{x\to 0}\left(\frac{-\frac{1}{x+1}}{6x+2}\right)\\\\
&=-e/2
\end{align}$$
A: With Taylor's expansion at order $\mathit 1$:
$$(1+x)^{\tfrac1x}=\mathrm e^\tfrac{\ln(1+x)}{x}=\mathrm e^{1-\tfrac{x}2+o(x)}=\mathrm e\Bigl(1-\frac x 2+o(x)\Bigr)$$
whence 
$$\frac{(1+x)^{\tfrac1x}-\mathrm e}x=\mathrm e\Bigl(-\frac12+o(1)\Bigr)\to-\frac{\mathrm e}2.$$
A: I wouldn't say it is "nasty" to differentiate $f(x)=(1+x)^{\frac{1}{x}}$. We have:
$$ f'(x) = f(x)\cdot\frac{d}{dx}\log(f(x)) = (1+x)^{\frac{1}{x}}\cdot\frac{d}{dx}\frac{\log(1+x)}{x} $$
and since in a neighbourhood of the origin $\log(1+x)=x-\frac{x^2}{2}+o(x^2)$ we have:
$$ \lim_{x\to 0^+} f'(x) = e\cdot\left(-\frac{1}{2}\right).$$
