Find$ \lim\limits_{n\rightarrow\infty}\frac{x_{n}}{n}$ We define $f:\mathbb{R}\rightarrow {R}$ by $f(x)=\frac{e^{x}-1}{x}$ if $x\ne0$ and $f(0)=1$ otherwise.
Let $(x_{n})_{n}$ the sequence defined by $x_{1}=\frac{1}{2}$,$x_{n+1}=\int_{0}^{x_{n}} f(t)dt$, for any $n\geq1$. Find $ \lim\limits_{n\rightarrow\infty}\frac{x_{n}}{n}$.
My progress is that I know that $\lim\limits_{t \rightarrow 0} \frac{e^{t}-1}{t}=1$ and the required limit, following the use of Cesaro-Stolz lemma, gives the limit of difference $x_{n+1}-x_{n}$. Also, using l'Hospital,$ \lim\limits_{n \rightarrow \infty} \frac{x_{n+1}}{x_{n}}=\lim\limits_{n \rightarrow \infty}  \frac{x_{n+1}}{e^{x_{n+1}}-1}$
 A: $f$ is an increasing, convex from above function (I'll prove it below for $x > 0$, which is the relevant part for this problem) and $f(x) > 0$ for $x \ge 0$. 
That means the tangent to it at any point is below the function. That means that the tangent at $(\frac{x_n}2,f(\frac{x_n}2))$ is below the function. 
That means the area enclosed by the $x$-axis, the two vertical lines $x=0$, $x=x_n$ and that tangent at $(\frac{x_n}2,f(\frac{x_n}2))$ is a subset of the area enclosed  by $x$-axis, the two vertical lines $x=0$, $x=x_n$ and $f(x)$.
The latter area has measure $\int_0^{x_n}f(t)dt = x_{n+1}$, the former is a trapezoid with height $x_n-0$ and middle line length $f(\frac{x_n}2)$, so has measure $x_nf(\frac{x_n}2)$.
That leads to the inequality
$$x_{n+1} = \int_0^{x_n}f(t)dt \ge x_nf(\frac{x_n}2)$$
With $f$ being increasing, it's easy to see by induction that 
$$x_{n+1} = f(\frac{x_n}2)x_n \ge f(\frac{x_1}2)x_n > x_n$$
as $f(\frac{x_1}2) > 1$.
This means $x_n \ge f(\frac{x_1}2)^{n-1}x_1$ and hence 
$$\lim_{n\to\infty}\frac{x_n}n \ge \lim_{n\to\infty}\frac{f(\frac{x_1}2)^{n-1}}nx_1 = \infty.$$

Proof of monotonicity and convexity:
$$f'(x)=\frac{e^x(x-1)+1}{x^2}$$
$$f''(x)=\frac{e^x(x^2-2x+2)-2}{x^3}$$
The limits at $x\to0$ for $f',f''$ can be easily evaluated with l'Hospital and shown to be postive. The denominators are also positive for $x > 0$. It remains to show that the enumarators of $f',f''$ are non-negative to show that $f',f'' \ge 0$ for $x > 0$.
That can be shown by considering the enumerators as functions of their own and looking at their value for $x=0$. Both enumerators have value $0$ there. 
Considering $g(x)=e^x(x-1)+1$, we have $g'(x)=xe^x$, which is positive for $x > 0$, so $g$ is increasing from $x \ge 0$ and since $g(0)=0$ we have $g(x) \ge 0$ for $x\ge 0$, which implies $f'(x) \ge 0$ for $x \ge 0$.
Similiarly, considering $h(x)=e^x(x^2-2x+2)-2$ we have $h'(x)=x^2e^x$ which together with $h(0)=0$ implies $h(x) \ge 0$ for $x \ge 0$ and hence $f''(x) \ge 0$ for $x \ge 0$
A: Here's what I got at the moment : 


*

*$x_n \to \infty$

*We have $x_{n+1}-x_n = \int_0^{x_n}(f(t)-1)dt$ > $(e-2)(x_n-1)$ since $f>e-1$ for $x>1$.

*As a result $x_{n+1}-x_n \to \infty$ and we can apply Cesaro lemma to conclude that $\frac{x_n}{n} \to \infty$
A: The function
$$
f(x) = {{e^{\,x}  - 1} \over x} = \sum\limits_{0\, \le \,n} {{{x^{\,n} } \over {\left( {n + 1} \right)!}}} 
$$
is strictly increasing and convex for all $x \in \mathbb R$.
Therefore
$$
\int_0^a {\left( {1 + {x \over 2}} \right)dx}  = a + {{a^{\,2} } \over 4} < \int_0^a {f(x)dx}  < {a \over 2}\left( {f(a) + f(0)} \right) = {{a + e^{\,a}  - 1} \over 2}
$$
that is
$$
x_{\,n} \left( {1 + {{x_{\,n} } \over 4}} \right) < x_{\,n + 1} \quad  \Rightarrow \quad {{x_{\,n} ^{\,2} } \over 4} < \Delta \,x_{\,n}  = x_{\,n + 1}  - x_{\,n} 
$$
