Finding a limit involving exponential function

Using the standard Taylor expansions (to low order) $\ln(1+u)=u-\frac{u^2}{2}+o(u^2)$ and $e^u=1+u+o(u)$ when $u\to0$.

Rewriting $$\begin{align} (1+x)^{\frac{1}{x}} &= \exp\left( \frac{1}{x}\ln(1+x)\right) = \exp\left( \frac{1}{x}(x-\frac{x^2}{2} + o(x^2))\right) = \exp\left( 1-\frac{x}{2} + o(x)\right) \\&= e\cdot \exp\left( -\frac{x}{2} + o(x)\right) = e\cdot \left( 1-\frac{x}{2} + o(x)\right) = e-e\frac{x}{2} + o(x) \end{align}$$ we get that $$ \frac{e-(1+x)^{\frac{1}{x}}}{x} = \frac{e\frac{x}{2} + o(x)}{x} =\frac{e}{2} + o(1)\xrightarrow[x\to0]{} \boxed{\frac{e}{2}}. $$

Note that we expanded $\ln(1+x)$ to order $x^2$, since we can "guess" the first order will get cancelled eventually by the $-e$ in the denominator. (Doing only the expansion to first order will only, basically, yield the limit $(1+x)^{1/x}\xrightarrow[x\to0]{}e$, and so we know we need better precision.)

In the second step, factoring the $e$ out of the product allows us to get $e^{-x/2+o(x)}$ instead of $e^{1-x/2+o(x)}$, which is required in order to use the expansion of $e^u$ (since this expansion holds when $u\to 0$, and while $\frac{x}{2}\to 0$, this is not the case for $1-\frac{x}{2}$.)

Hint. Expand the function $f(x):=(1+x)^{\frac{1}{x}}=\exp\left(\frac{\ln(1+x)}{x}\right)$ at $0$: $$f(x)=\exp\left(\frac{x-\frac{x^2}{2}+o(x^2)}{x}\right)=\exp\left(1-\frac{x}{2}+o(x)\right)=e\cdot \exp\left(-\frac{x}{2}+o(x)\right).$$ Can you take it from here? At the end you should find that the limit is $e/2$.

Put $$f (x)=(1+x)^\frac 1x $$ for $x\ne 0$ and $f (0)=e $.

then your limit is $$-\lim_0\frac {f (x)-f (0)}{x-0} $$ or $$-\lim_0 f'(x ).$$ with

$f'(x)=f (x)(-\frac {1}{x^2}\ln (1+x)+\frac {1}{x (1+x)}) $

$=f (x)(\frac {-x+x^2/2 (1+\epsilon (x))}{x^2}+\frac {1}{x}-\frac {1}{1+x}) $

from this, the limit is $$-e(\frac {1}{2}-1)=\frac {e}{2} $$