Find a limit for $\frac{({1+x})^\frac{1}{x} - e}{x}$ as x tends to 0 I have done it thus far:
$$\lim_{x \to 0}\frac{{(x+1)}^\frac{1}{x}-e}{x} = \bigg[\frac{0}{0}\bigg] = \frac{((x+1)^\frac{1}{x}-e)'}{x'}=({x+1})^\frac{1}{x} \cdot \left(\frac{\ln(x+1)}{x}\right)' = \\({x+1})^\frac{1}{x} \cdot \frac{\left(\frac{x}{x+1}-(\ln(x+1)\right)}{x^2} = ({x+1})^\frac{1}{x} \cdot \left(\frac{\frac{1}{x+1}}{x} - \frac{\ln(x+1)}{x^2}\right) $$
I don't know what to do next.
Also could someone elaborate as to why when I have to find a derivative for $\frac{\ln(x+1)}{x}$ I need to use the quotient rule, but when I first derived the fraction that is given I could derive numerator and denominator separately? 
 A: $$\begin{align}
\lim_{x\to0}\left(\frac{\frac{x}{x+1}-\ln{(x+1)}}{x^2}\right)
&=\lim_{x\to0}\left(\frac{x-(x+1)(x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4))}{x^2(x+1)}\right)\\
&=\lim_{x\to0}\left(\frac{-x^2+\frac{x^2}{2}+\frac{x^3}{2}-\frac{x^3}{3}+O(x^4)}{x^3+x^2}\right)\\
&=\lim_{x\to0}\left(\frac{-\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)}{x^3+x^2}\right)\\
&=\lim_{x\to0}\left(\frac{-x+\frac{x^2}{2}+O(x^3)}{3x^2+2x}\right)\\
&=\lim_{x\to0}\left(\frac{-1+x+O(x^2)}{6x+2}\right)\\
&=-\frac12\\
\end{align}$$
Hence the initial limit becomes
$$\overbrace{\lim_{x\to0}\left((x+1)^\frac1x\right)}^{\large{\to\, e}}\cdot\overbrace{\lim_{x\to0}\left(\frac{\frac{x}{x+1}-\ln{(x+1)}}{x^2}\right)}^{\large{\to\, -\frac12}}=-\frac{e}{2}$$
A: So, we apply L'Hôpital's rule,  and get $((1+x)^{\dfrac 1x})'=(e^{\ln(1+x)^{\dfrac 1x}})'=(e^{\ln(1+x)^{\dfrac1x}}\cdot(\dfrac 1x\ln(1+x))'=(1+x)^{\dfrac 1x}\cdot (-\dfrac 1{x^2}\ln(1+x)+\dfrac1x\dfrac 1{1+x})$.
Now the first term converges to $e$;  and the quantity in parentheses converges easily, upon a couple more applications of L'Hôpital to $-\dfrac 12$.
So we get $-\dfrac e2$.
A: You want to compute the derivative of
$$
f(x)=(1+x)^{1/x}
$$
Consider $\log f(x)=\dfrac{\log(1+x)}{x}$ and
$$
\frac{f'(x)}{f(x)}=\frac{\dfrac{x}{1+x}-\log(1+x)}{x^2}
$$
so you have
$$
f'(x)=f(x)\frac{x-(1+x)\log(1+x)}{x^2(1+x)}
$$
Note also that $\lim\limits_{x\to0}\dfrac{f(x)}{1+x}=e$, so you need to compute
$$
\lim_{x\to0}\frac{x-(1+x)\log(1+x)}{x^2}
$$
With a further application of l'Hôpital,
$$
\lim_{x\to0}\frac{1-1-\log(1+x)}{2x}=-\frac{1}{2}\lim_{x\to0}\frac{\log(1+x)}{x}=-\frac{1}{2}
$$
Reinserting the factor $e$, the sought limit is $-e/2$.
A: With binomial expansion 
$$\begin{align}
(1+x)^{\frac1x}-e 
&=1+\frac1x\frac{x}{1!}+\frac1x(\frac1x-1)\frac{x^2}{2!}+\frac1x(\frac1x-1)(\frac1x-2)\frac{x^3}{3!}+\cdots-\left(1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\cdots\right)\\
\frac{(1+x)^{\frac1x}-e }{x}&=
-\dfrac{1}{2!}-\dfrac{3-2x}{3!}-\dfrac{6-11x+6x^2}{4!}+\cdots\\
&\to-\dfrac{1}{2!}\left(1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots\right)\hspace{1cm}\text{as}\hspace{0.3cm} x\to0.\\
&=-\dfrac12e
\end{align}$$
