# 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 $x \to \infty$? 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. Aug 2 '17 at 23:50
• @Somos Because it is proven it works Aug 2 '17 at 23:51
• 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). Aug 3 '17 at 2:41
• L'Hôpital only applies when your limit is either $\dfrac00$ or $\dfrac{∞}{∞}$. Sep 19 at 15:15

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 $\lim f(x)/g(x) = L.$

L'Hospital's Rule has a lot of hypotheses. You seem to be ignoring the hypotheses on the limits of the numerator and denominator of the original ratio. Let's check a statement of the Rule.

From Calculus: Early Transcendentals, 4th ed. by Rogawski, Adams, Franzosa (p. 250):

Assume $$f$$ and $$g$$ are differentiable in an interval $$(b,\infty)$$ and that $$g'(x) \neq 0$$ for $$x > b$$. If $$\lim_{x \rightarrow \infty} f(x)$$ and $$\lim_{x \rightarrow \infty} g(x)$$ exist and either both are zero or both are infinite, then $$\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)}$$ provided that the limit on the right exists. A similar result holds for $$x \rightarrow -\infty$$.

(This quote contains an astoundingly common error. As Rogawski explains on p. 72, infinite limits do not exist. A better phrasing of the second sentence is "If it is the case that $$\lim_{x \rightarrow \infty} f(x)$$ and $$\lim_{x \rightarrow \infty} g(x)$$ both exist and are zero or it is the case that both limits are infinite [which in Rogawski includes either sign], then ...". This error is common among those who have studied real analysis because in that setting one works in the extended reals, so $$\infty$$ and $$-\infty$$ are points in the working set of numbers and can be the value of a limit. In (just) the reals, this is not the case.)

Applying this to your example, $$f(x) = \sin x$$ and $$g(x) = x$$. While it is the case that $$\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} x = 0 \text{,}$$ we find that
$$\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} \sin x$$ does not exist. (The sine function takes every interval $$(b,\infty)$$ to $$[-1,1]$$, so the limit fails to exist. One can arrive at the same conclusion by studying $$\lim_{x \rightarrow 0^+} \sin(1/x)$$, but with the added advantage that the graph fits on a finite piece of paper.)

So, no, your example is not a special case. L'Hospital's Rule already constrains to which limit expressions it can be applied, excluding the example you give.

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. Aug 3 '17 at 0:49
• L'Hopital can be applied whenever the denominator $\to \infty.$
– zhw.
Aug 3 '17 at 0:51
• @zhw. How do you figure? By OP's example, this should be false. Aug 3 '17 at 2:15
• 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
• @zhw. : Then your comment is not an improvement of this Answer in the context of the Question. $f'/g' = \cos(x)$ does not have a limit as $x \rightarrow \infty$. For careful statements of L'Hopital's Rule, your first comment ignores necessary hypotheses on the existence of various limits, claiming, for instance, $\lim_{x \rightarrow \infty} \frac{x \sin x}{x}$ exists, which is false (although your second comment does correct for this particular example). Sep 19 at 14:42