# A question about limits and derivatives [duplicate]

$f:(0,\infty)\to \mathbb{R}$ is differentiable such that $f'(x)\to l$ as $x\to \infty$. Prove that $\frac{f(x)}{x}\to l$ as $x\to \infty$.

## marked as duplicate by Mark S., Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 10 '16 at 19:04

• Hospital?${}{}{}$ – David Mitra Dec 10 '16 at 18:45
• 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. – Lav Kumar Dec 10 '16 at 18:47
• @LavKumar No, LHR does not require that $f\to \infty$. – Mark Viola Dec 10 '16 at 18:48
• This might help. – David Mitra Dec 10 '16 at 18:49

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!

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