Proof of $\lim\limits_{x\to 0} \frac{e - (1+x)^\frac{1}{x}}{x}=\frac{e}{2}$ [duplicate]
I want to prove $$\lim_{x\to 0} \frac{e - (1+x)^\frac{1}{x}}{x}=\frac{e}{2}$$ without using L'Hôpital's rule, and Taylor Series.
Solution 1:
Hint
First, consider $A=(1+x)^\frac{1}{x}$. Take the logarithms of both sides so $$\log(A)=\frac{1}{x} \log(1+x)$$ Use the Taylor series of $\log(1+x)$; multiply the result by $\frac{1}{x}$; take the exponential; use the Taylor series and so on.
I am sure that you can take from here.