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

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.