How to evaluate $\lim_{x\to0} \frac{(1+x)^{\frac1x} - e}{x}$ with Taylor series

Question

I'm stuck in solving this limit $$ \lim_{x\to0} \frac{(1+x)^{\frac1x} - e}{x}. $$

Here I can must use Taylor expansion. My idea is to obtain the form $e^y-1$ on the numerator and then use Taylor expansion on it stopping on the first grade but I continue to fail. I also tried to stop on the second grade but no results. Note that I started using Taylor a few hours ago so I'm not so familiar. Can you help me? Thank you in advance. (sorry for my english)

Answer 1

$$(1+x)^{\frac1x}=\exp(\log((1+x)^{\frac1x}))=\exp(\frac1x\log(1+x))$$ $$(1+x)^{\frac1x}-e=\exp(\frac1x\log(1+x))-e=e\left(\exp\left(\frac1x\log(1+x)-1\right)-1\right)$$ And use $\exp(f(x))-1\approx f(x)$ when $f(x)\to 0$.

Answer 2

Let us consider first $$A=(1+x)^{\frac 1x}$$ So, taking logarithms, $$\log(A)={\frac 1x}\log(1+x)={\frac 1x}\Big(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right)\Big)=1-\frac{x}{2}+\frac{x^2}{3}+\cdots$$ So, $$A=e^{1-\frac{x}{2}+\frac{x^2}{3}+\cdots}=e \times e^{-\frac{x}{2}+\frac{x^2}{3}+\cdots}$$ For small $y$, $$e^y=1+y+\frac{y^2}{2}+\frac{y^3}{6}+O\left(y^4\right)$$ Now, replace in $A$, $y$ by $(-\frac{x}{2}+\frac{x^2}{3})$ and obtain $$A=e\Big(1-\frac{x}{2}+\frac{11 x^2}{24}+\cdots\Big)$$

I am sure that you can take from here.