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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

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