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## Convex function & limit

Suppose $f: [0,\infty) \to \mathbb{R}$ is convex and differentiable with $\lim\limits_{x\to \infty} \frac{f(x)}{x}=\ell \in \mathbb{R}$. Prove that $\lim\limits_{x \to \infty} f^\prime (x)=\ell$.
Solution: Since $f$ is convex, its derivative is increasing, therefore, $\exists \lim_{n\to \infty}f'(x)=\ell'$.
By the definition of the limit we have that for any $\varepsilon >0$ there exists $x_0 >0$ such that for any $x >x_0$ we have $(\ell-\varepsilon)x < f(x) \ell$ there exists $x_0>0$ and $\varepsilon>0$ such that for any $x>x_0$ we have $f^\prime(x) > \ell+2\varepsilon$. Therefore $g(x)=f(x)-(\ell+\varepsilon)x$ is strictly increasing, but is bounded above by 0 and $g'(x)>\varepsilon ,\ \forall x \geq x_0$ large enough. This means that $g(x)> \varepsilon x +g(x_0),\ \forall x \geq x_0$, which means that $\lim_{n\to \infty} g(x)=\infty$. This contradicts $g(x)<0,\ x \geq x_0$. The case $\ell^\prime < \ell$ is solved in a similar way