If $\lim_{x\to\infty}\frac{\sin x}{x} = 0$ is zero, why does it not work with L'hospital's way?

I just got a simple question regarding the use of L'Hopitals method for finding limits. Usually L'Hopitals method can be used to find limits like $$\lim_{x\to0}\frac{\sin x}{x} = \lim_{x\to 0}\dfrac{\dfrac{d}{dx} \sin x}{\dfrac{d}{dx} x} = \lim_{x\to 0}\cos x$$

Here if we plug $0$, we can find the limit of the original function $\dfrac{\sin x}{x}$ at $0$ using the $\cos x$ function. Put $0$ in, and you will get $1$, which is correct. However, if we replace $x$ with $\infty$, we don't get the right limit.

$$\cos x$$ $$\cos(\infty)$$

Which is not right for the limit of the original function, as $$\lim_{x\to\infty}\frac{\sin x}{x} = 0$$ Using the new function which we get via L'Hopital's method does not help get that. Is this like a special case? In what cases could then L'Hopital's way not work?

• Why do you think that L'Hospital's rule works for $x \to \infty$? – Somos Aug 2 '17 at 23:45
• $\sin x$ has no limit as $x \to \infty$ (and also does not approach either $\infty$ or $-\infty$) so the limit is not in one of the valid indeterminate forms for applying l'Hopital. – Daniel Schepler Aug 2 '17 at 23:50
• @Somos Because it is proven it works – Peyton Aug 2 '17 at 23:51
• @Peyton I have read the Wikipedia article more carefully and indeed the point $c$ can be infinity, but there are other requirements as well on the limits of $f(x)$ and $g(x)$. – Somos Aug 2 '17 at 23:56
• The main issue with L'Hospital's Rule is that most often one does not pay attention to all the hypotheses under which it works. Perhaps the rule is thought as a thumb rule of differentiate and plug instead of a genuine mathematical theorem. One of the hypotheses of the rule is that the limit of the $f'/g'$ (expression obtained after differentiation of numerator and denominator) should have a limit (finitely or infinitely). – Paramanand Singh Aug 3 '17 at 2:41

L'Hopital has the form if $A,$ then $B.$ It doesn't say $A$ iff $B.$ Thus the limit of $f/g$ can equal $L$ (B) even if the limit of $f'/g'$ doesn't exist.
Incidentally, L'Hopital works in cases of $\text { ? }/\infty.$ In other words, if $\lim g(x) = \infty$ and $\lim f'(x)/g'(x) = L,$ then $\lim f(x)/g(x) = L.$
You can directly compute the limit of $\frac{sin(x)}{x}$ as $x \to \infty$, which is 0 and not an indeterminate form - this, L'Hopital's doesn't apply.
• Just because you can calculate the limit in another way doesn't automatically mean l'Hopital doesn't apply. e.g. you can calculate $\lim_{x \to \infty} \frac{x}{x}$ easily enough without l'Hopital, but l'Hopital also works on it. – Daniel Schepler Aug 3 '17 at 0:49
• L'Hopital can be applied whenever the denominator $\to \infty.$ – zhw. Aug 3 '17 at 0:51
• By "can be applied" I mean that if $g\to \infty,$ and if $f'/g'\to L,$ then $f/g\to L.$ – zhw. Aug 3 '17 at 3:17