Prove that $\lim_{x \to \infty} \frac{f(x)}{x}$ exists This is a problem that has been asked in an old analysis 1 exam. I can't find it anywhere else online.
Let $f$ be twice differentiable on $(1,\infty)$ such that $f\geq 0$ and $f''\leq 0$. Show that $\lim_{x \to \infty } \frac{f(x)}{x} $ exists.
I think you have to show that $\frac{f(x)}{x}$ is decreasing using the MVT, but I don't know how to do it.
 A: *

*$f'$ is decreasing because $f''\le 0.$ If $f'(x_0)<0$ for some $x_0>1$ then when $x>x_0$ we have $f(x)=f(x_0)+\int_{x_0}^x f'(y)dy\le f(x_0)-(x-x_0)\cdot|f'(x_0)|\to -\infty$ as $x\to\infty,$ contrary to $f\ge 0.$
Therefore $f'\ge 0.$ But $f'$ is decreasing, so $f'(x)$ has a limit $L\ge 0$ as $x\to\infty.$


*Given any $e>0,$ take $x_1>1$ such that $y\ge x_1\implies L\le f'(y)\le L+e.$ Now if $x>x_1$ then $$f(x)=f(x_1)+\int_{x_1}^x f'(y)dy.$$
Upper and lower bounds for the last expression (above) are $f(x_1)+(L+e)(x-x_1)$ and  $f(x_1)+(L)(x-x_1).$
Dividing these bounds by $x,$ and letting $x\to\infty,$ we have $$L+e\ge\lim_{z\to\infty}\sup_{x>z}\frac {f(x)}{x}\ge\lim_{z\to\infty}\inf_{x>z}\frac {f(x)}{x}\ge L.$$ Since this holds for every $e>0,$ therefore $\lim_{x\to\infty}\dfrac {f(x)}{x}=L.$
A: Since $f(x)\ge 0$ and $f''(x)\le $, for all $x>1$, then $f'(x)\ge 0$, for all $x\ge 0$. Otherwise, if $f'(x_0)<0$, for some $x_0>1$, then $f'(x)\le f'(x_0)$, for all $x>x_0$ and hence
$$
f(x)= f(x_0)+\int_{x_0}^x f'(t)\,dt\le f(x_0)+(x-x_0)f'(x_0)\to -\infty
$$
and consequently, $f(x)\to -\infty$. Contradiction.
So $f'(x)\ge 0$ and $f'$ is decreasing. This implies that $\lim_{x\to\infty} f'(x)$ exists is $[0,\infty)$. Let $L=\lim_{x\to\infty} f'(x)$.
Let $x_0>1$, then for every $x>1$ we have that
$$
f(x)=f(x_0)+\int_{x_0}^x f'(t)\,dt\quad\Longrightarrow\quad
\frac{f(x)}{x}=\frac{f(x_0)}{x}+\frac{1}{x}\int_{x_0}^x f'(t)\,dt.
$$
Clearly, $\frac{f(x_0)}{x}\to 0$, whereas
$$
\frac{1}{x}\int_{x_0}^x f'(t)\,dt=\frac{1}{x}\int_{x_0}^x \big(f'(t)+1\big)\,dt-\frac{x-x_0}{x}\to L+1-1=L,
$$
since $\int_{x_0}^x \big(f'(t)+1\big)\,dt\ge x-x_0\to \infty$, and using L'Hopital,
$$
\lim_{x\to\infty}\frac{\int_{x_0}^x \big(f'(t)+1\big)\,dt}{x}=
\lim_{x\to\infty}\frac{f'(x)+1}{1}=L+1.
$$
