Show that $\lim_{n\rightarrow\infty} \int_{n}^{n+1}f\left(\frac{x}{(n!)^{\frac{1}{n}}}\right)dx=f(e)$ Let $f:(0,\infty)\rightarrow\mathbb{R}$ be a contiuous function. Show that $\lim_{n\rightarrow\infty} \int_{n}^{n+1}f\left(\frac{x}{(n!)^{\frac{1}{n}}}\right)dx=f(e)$. Because $\ln(n!)^{\frac{1}{n}}=\frac{\ln2+\ln3+...+\ln n}{n}$, by applying Stolz-Cesaro lemma I get that $\lim_{n\rightarrow\infty}(n!)^{\frac{1}{n}}=\lim_{n\rightarrow\infty}\ln(n+1)-\ln(n)=\lim_{n\rightarrow\infty}\ln(\frac{n+1}{n})=0$ and so $\lim_{n\rightarrow\infty}(n!)^{\frac{1}{n}}=e$. I tried to apply from here on different strategies to solve it but the fact I have $(\frac{x}{(n!)^{\frac{1}{n}}})$ under the integral sign prevented any success. Any help, please?
 A: Note that 
$$\int_n^{n+1} f \left( \frac{x}{\left(n!\right)^{1/n} }\right) \mathrm{dx} = \int_0^1  f \left( \frac{x+n}{\left(n!\right)^{1/n} }\right) \mathrm{dx} $$
Consider the sequence of functions $(f_n : [0,1] \rightarrow \mathbb{R})_{n \in \mathbb{N}^*}$ defined by
$$\forall n \in \mathbb{N}^*, \forall x \in [0,1], \quad f_n(x)= f \left( \frac{x+n}{\left(n!\right)^{1/n} }\right)$$
By Stirling formula, you have
$$\left(n!\right)^{1/n} \sim \frac{n}{e}$$
So for all $x \in [0,1]$,
$$\lim_{n \rightarrow +\infty} \frac{x+n}{\left(n!\right)^{1/n}} = e, \quad \text{so } \lim_{n \rightarrow +\infty} f_n(x) = f(e) $$
So the sequence $(f_n)$ converges simply to the constant function equal to $f(e)$. Moreover, for all $x \in [0,1]$, 
$$0 \leq \frac{x+n}{\left(n!\right)^{1/n}} \leq \frac{n+1}{\left(n!\right)^{1/n}} $$
and the right term of this inequality is bounded by a certain number $M$ (as a convergent sequence), so $f_n(x)$ is bounded by the sup of $f$ on $[0,M]$, which is integrable as a constant function on $[0,1]$. By dominated convergence theorem, you deduce that 
$$\lim_{n \rightarrow +\infty} \int_0^1 f_n(x) \mathrm{dx} = \int_0^1 f(e) \mathrm{dx} = f(e)$$
A: By the MVT of integration, the integral 
$$
I(n) = f\left(\frac \xi {(n!)^{1/n}}\right), \xi \in [n, n+1]. 
$$
Since
$$
\lim_n \frac n{(n!)^{1/n}} = \lim_n \frac n {(\sqrt {2\pi n })^{1/n} (n/\mathrm e)} = \mathrm e
$$
by Stirling approximation, and
$$
\lim_n \frac {n+1} {(n!)^{1/n}} = \lim_n \frac n{(n!)^{1/n}} \times \frac {n+1}n = \mathrm e, 
$$
then by squeezing theorem
$$
\lim_n \frac \xi{(n!)^{1/n}} = \mathrm e. 
$$
By the continuity of $f$, 
$$
\lim_n I(n) = f \left(\lim_n \frac \xi{(n!)^{1/n}}\right) = f (\mathrm e). 
$$
