$f:(0,\infty)\to \mathbb{R}$ is differentiable such that $f'(x)\to l$ as $x\to \infty$. Prove that $ \frac{f(x)}{x}\to l $ as $x\to \infty$.
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Hospital?${}{}{}$ – David Mitra Dec 10 '16 at 18:45
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I am looking for a more analytical proof. Besides how do we apply Hospital's. We need to first show that f(x) tends to infinity as x does. – NewB Dec 10 '16 at 18:47
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@LavKumar No, LHR does not require that $f\to \infty$. – Mark Viola Dec 10 '16 at 18:48
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2This might help. – David Mitra Dec 10 '16 at 18:49
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Since $f'(x)\to \ell$, then for all $\epsilon>0$, there exists a number $x_0$ such that $\ell-\epsilon< f'(x)<\ell +\epsilon$ whenever $x>x_0$.
From the mean value theorem, we have
$$f(x)=f(x_0)+f'(\xi)(x-x_0)$$
for some $\xi \in (x_0,x)$. But then we can write for $x>x_0$
$$\frac{f(x_0)}{x}+\left(1-\frac{x_0}{x}\right)(\ell -\epsilon)<\frac{f(x)}{x}<\frac{f(x_0)}{x}+\left(1-\frac{x_0}{x}\right)(\ell +\epsilon)$$
Letting $x\to \infty$ we see that for all $\epsilon>0$ we have
$$\ell -\epsilon<\lim_{x\to \infty}\frac{f(x)}{x}<\ell +\epsilon$$
and hence $\lim_{x\to \infty}\frac{f(x)}{x}=\ell$ as was to be shown!
Mark Viola
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