How fast does (1 + x)^(1/x) reach e?

Idea. The limit tells us the *size of the error* when we replace $(1+x)^{1/x}$ by $e$. We need a more precise expansion than the one that only proves convergence.

Step 1: the exponent. As $x\to0$, $\ln(1+x)=x-\dfrac{x^2}2+o(x^2)$, so
$$\frac{\ln(1+x)}{x}=1-\frac x2+o(x)=1+h,\qquad h=-\frac x2+o(x).$$

Step 2: the exponential. Since $h\to0$, $e^h=1+h+o(h)=1-\dfrac x2+o(x)$. Hence
$$(1+x)^{1/x}=e\cdot e^h=e-\frac{e}{2}\,x+o(x).$$

Step 3: conclusion.
$$\frac{(1+x)^{1/x}-e}{x}=-\frac e2+o(1)\longrightarrow-\frac e2.$$

Numerical check. $-\frac e2\approx-1.3591$, and the quotient equals $-1.2454$ for $x=0.1$, $-1.3468$ for $x=0.01$, $-1.3579$ for $x=0.001$.

Meaning. For small $x$, $(1+x)^{1/x}\approx e\left(1-\frac x2\right)$: the convergence to $e$ is only *linear* in $x$.