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? 

With binomial expansion 
$$\begin{align}
(1+x)^{\frac1x}-e 
&=1+\frac1x\frac{x}{1!}+\frac1x(\frac1x-1)\frac{x^2}{2!}+\frac1x(\frac1x-1)(\frac1x-2)\frac{x^3}{3!}+\cdots-\left(1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\cdots\right)\\
\frac{(1+x)^{\frac1x}-e }{x}&=
-\dfrac{1}{2!}-\dfrac{3-2x}{3!}-\dfrac{6-11x+6x^2}{4!}+\cdots\\
&\to-\dfrac{1}{2!}\left(1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots\right)\hspace{1cm}\text{as}\hspace{0.3cm} x\to0.\\
&=-\dfrac12e
\end{align}$$
