How to solve this limit: $\lim\limits_{x\to0}\frac{(1+x)^{1/x}-e}x$? I tried to evaluate $$\lim_{x\to0}\frac{(1+x)^{1/x}-e}x$$but all my efforts were in vain . L'Hospital rule becomes very complicated.
 A: You want to find 
$L = \lim_{x\to0}\dfrac{(1+x)^{1/x}-e}x$.
Let $f(x) = (1+x)^{1/x}$.
Since $\lim_{x\to0} f(x) = e$,
by the definition of derivative,
$L = f'(0)$.
Applying the chain rule
$(f(g(x))' = g'(x)f'(g(x))$
in the form
$(e^{g(x)})' = g'(x) e^{g(x)}$
very carefully,
$\begin{align}
f'(x)
&= ((1+x)^{1/x})'\\
&= (\exp((1/x)\ln(1+x))'\\
&= ((1/x)\ln(1+x))'(\exp((1/x)\ln(1+x))\quad\quad( (e^x)' = e^x)\\
&= ((1/x)/(1+x)-\ln(1+x)/(x^2))'(1+x)^{1/x}\quad(\text{Product rule})\\
&= \left(\dfrac{1}{x(1+x)}-\dfrac{\ln(1+x)}{x^2}\right)(1+x)^{1/x}\\
&= \left(\dfrac{x-(1+x)\ln(1+x)}{x^2(1+x)}\right)(1+x)^{1/x}\\
\end{align}
$
Since
$\lim_{x \to 0+} (1+x)^{1/x} = e$,
we only need to evaluate the
left-hand term as $x \to 0+$.
For this, L'Hospital's rule rules.
$\begin{align}
\lim_{x \to 0+} \dfrac{x-(1+x)\ln(1+x)}{x^2(1+x)}
&=\lim_{x \to 0+}\dfrac{1-((1+x)/(1+x)+\ln(1+x)}{2x(1+x)+x^2}\\
&=\lim_{x \to 0+}\dfrac{-\ln(1+x)}{2x+x^2}\\
&=\lim_{x \to 0+}\dfrac{-1/(1+x)}{2+2x}\quad\quad\text{Applying Hoppy again}\\
&= -\dfrac{1}{2}
\end{align}
$
So $f'(0) = -\dfrac{e}{2}$.
A: Write, for $x > 0$,
$$\begin{align*}
\frac{(1+x)^{1/x}-e}x &= \frac{e}{x}\left( e^{\large\frac{1}{x}\ln(1+x)-1}-1 \right) \\
&=\frac{e}{x}\left( e^{\large\frac{x-\frac{x^2}{2}+o(x^2)}{x}-1}-1 \right) \\
&=\frac{e}{x}\left( e^{\large-\frac{x}{2}+o(x)}-1 \right) \\
&=\frac{e}{x}\left( 1-\frac{x}{2}+o(x)-1 \right) \\
&=e\left( -\frac{1}{2}+o(1) \right) \xrightarrow[x\to0^+]{}-\frac{e}{2}
\end{align*}
$$
A: By definition, $(1+x)^{1/x}=\exp(\frac1x\ln(1+x))$. Using Taylor series
$$\ln(1+x)=x-\frac12x^2+O(x^3) $$
$$\exp(x)=1+x+\frac12x^2+O(x^3)$$
we find
$$\begin{align}(1+x)^{1/x}-e&=e\cdot(\exp(\frac1x\ln(1+x)-1)-1)\\&=e\cdot(\exp(-1+1-\frac12x+O(x^2))-1) \\&= e\cdot(\exp(-\frac12x)-1+O(x^2))\end{align}$$
Then $ \frac{\exp(-\frac12x)-1}{x}$ tends (by definition) to the derivative of $x\mapsto \exp(-\frac12x)$ at $x=0$.
A: Apply L'Hospital's rule for $\frac00$ form
$$\lim_{x\to 0}\frac{(1+x)^{1/x}-e}{x}$$
$$=\lim_{x\to 0}\frac{\frac{d}{dx}(1+x)^{1/x}-\frac{d}{dx}e}{\frac{d}{dx}x}$$
$$=\lim_{x\to 0}(1+x)^{1/x}\cdot \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$$
$$=\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$$
apply L'Hospital's rule for $\frac00$ form of second limit as follows
$$=\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{\frac{d}{dx}\frac{x}{1+x}-\frac{d}{dx}\ln(1+x)}{\frac{d}{dx}x^2}$$
$$=\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{\frac{1}{(1+x)^2}-\frac{1}{(1+x)}}{2x}$$
$$=\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{-x}{2x(x+1)^2}$$
$$=-\frac12\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{1}{(x+1)^2}$$
$$=-\frac12(e)\cdot (1)$$
$$=-\frac e2$$
