Prove that $\lim _{x\to \infty \:}(1+\frac{x^x}{x!})^{\frac{1}{x}} = e$ Using a graphing calculator, it seems that $\lim _{x\to \infty \:}(1+\frac{x^x}{x!})^{\frac{1}{x}} = e$. How can this be proven?

Since
$n! > (n/e)^n$,
for integer $x$,
$1+\frac{x^x}{x!}
<1+\frac{x^x}{(x/e)^x}
=1+e^x
$.
Therefore,
for integer $x$,
$(1+\frac{x^x}{x!})^{\frac{1}{x}}
< (1+e^x)^{1/x}
=e (1+e^{-x})^{1/x}
=e (1+\frac{e^{-x}}{x})
=e +e\frac{e^{-x}}{x}
$.
Since
$n! < (n/e)^{n+1}$,
for integer $x$,
$1+\frac{x^x}{x!}
\gt 1+\frac{x^x}{(x/e)^{x+1}}
=1+\frac{e^{x+1}}{x}
$.
Therefore,
for integer $x$,
$(1+\frac{x^x}{x!})^{\frac{1}{x}}
> (1+\frac{e^{x+1}}{x})^{1/x}
= e (e^{-x}+\frac{e}{x})^{1/x}
= e \left(\frac{e}{x}\right)^{1/x}(1+e^{-x}\frac{x}{e})^{1/x}
\gt e +\left(\frac{e}{x}\right)^{1/x}
$.
Therefore,
for integer $x$,
$e \left(\frac{e}{x}\right)^{1/x}
<(1+\frac{x^x}{x!})^{\frac{1}{x}}
<e +e\frac{e^{-x}}{x}
$
and the limits of both sides
are $e$.
