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$f:(0,\infty)\to \mathbb{R}$ 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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marked as duplicate by Mark S., Namaste calculus Dec 10 '16 at 19:04

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  • $\begingroup$ Hospital?${}{}{}$ $\endgroup$ – David Mitra Dec 10 '16 at 18:45
  • $\begingroup$ 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. $\endgroup$ – Lav Kumar Dec 10 '16 at 18:47
  • $\begingroup$ @LavKumar No, LHR does not require that $f\to \infty$. $\endgroup$ – Mark Viola Dec 10 '16 at 18:48
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    $\begingroup$ This might help. $\endgroup$ – 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!

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  • $\begingroup$ Thank you. This really helped ! $\endgroup$ – Lav Kumar Dec 10 '16 at 19:02
  • $\begingroup$ You're welcome! My pleasure. $\endgroup$ – Mark Viola Dec 10 '16 at 19:04

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