How can prove this question?
Let $f: [0.1]\to\mathbb{R}$ be a convex function of class $C^1$, with f'(0)> 0.
Prove that
$\lim_{x \to +\infty}f(x)=+\infty$
And that the limit exists finite
$\lim_{x \to +\infty}\frac{x}{f(x)}$
How can prove this question?
Let $f: [0.1]\to\mathbb{R}$ be a convex function of class $C^1$, with f'(0)> 0.
Prove that
$\lim_{x \to +\infty}f(x)=+\infty$
And that the limit exists finite
$\lim_{x \to +\infty}\frac{x}{f(x)}$
A standard result is that a $C^1$ convex function on $\mathbb R$ (or any open interval) has an increasing derivative. So f(x)=\int_0^xf'(t)dt+f(0)\geq\int_0^xf'(0)dt+f(0)=xf'(0)+f(0), and we get the result since f'(0)>0.
For the second question, put $g(x)=\frac{f(x)}x$ for $x>0$. Then g'(x)=\frac{f'(x)x-f(x)}{x^2}. Now fix $x_0>0$ such that $f(x_0)>0$. Then for $x\geq x_0$ g'(x)=\frac{f'(x)(x-x_0)+f'(x)x_0-f(x)}{x^2}=\frac{f'(x)(x-x_0)+f'(x)x_0-\int_{x_0}^xf'(t)dt+f(x_0)}{x^2}, and since f'(x)(x-x_0)-\int_{x_0}^xf'(t)dt\geq 0 and f'(x)x_0+f(x_0)\geq 0, we get that $h$ is increasing on $[x_0,+\infty[$ so $x\mapsto \frac x{f(x)}$ is decreasing on $[x_0,+\infty[$. Therefore, the limit $\lim_{x\to +\infty}\frac x{f(x)}$ exists and is finite since $\frac x{f(x)}> 0$ for $x$ large enough.