Limit of a function without LH

Question

Evaluate the limit : ${((1+x)^{1/x} -e})/x$ when x tends to zero.

This can be solved using L-Hopital rule , however I was wondering if there is any other method to do it, like any series formula or any other thing?

Answer 1

Considering $$y=\frac{(1+x)^{\frac 1x} -e}x$$ compose Taylor series $$a=(1+x)^{\frac 1x}\implies \log(a)=\frac 1x\log(1+x)$$ $$\log(a)=\frac 1x\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right) \right)=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$ $$a=e^{\log(a)}=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$ Just finish : you will get not only the limit but also how it is approached.