Suppose the equation $\displaystyle \frac{e^x}{x}=\int_n^{n+1}f(t)\,dt$ as $f(t)=\frac{e^t}{t}$ and $n\in \mathbb{N} \setminus{0}$.
How to prove that:
- The equation above has a unique solution $U_n$ on $[n,n+1]$.
- The $\displaystyle \lim_{n \to +\infty} \frac{U_n}{n}=1$.
Take $F(x)=\int_0^x f(t)\, dt$
Then $\int_n^{n+1}f(t)\,dt = F(n+1)-F(n)$ and $f=F'$
So by the mean value theorem, you have $U_n \in [n,n+1]$ so that $F'(U_n)=\cfrac{F(n+1)-F(n)}{(n+1)-n}$
$n \le U_n \le n+1$
$1 \le \cfrac{U_n}{n} \le 1+\cfrac{1}{n} \underset{n\to +\infty}{\longrightarrow} 1$
So $\cfrac{U_n}{n}\underset{n\to +\infty}{\longrightarrow} 1$