Evaluate $\lim_{x \to \infty} \sqrt{x} \sin(\frac{1}{x})$

I am trying to evaluate $\lim_{x \to \infty} \sqrt{x} \sin(\frac{1}{x})$. I have been taught that L'Hopital's Rule is only valid for fractions $\frac{f(x)}{g(x)}$ which have the form where $f(x) = g(x) = 0$ or where $g(x) = \pm \infty$ and $f(x)$ is anything.

Right away I notice that this limit evaluates to $\infty \cdot 0$. I need to put this in either the $\frac{0}{0}$ or the $\frac{\text{anything}}{\pm \infty}$ form to use L'Hopital's rule.

So I write $\sqrt{x} \sin(\frac{1}{x}) = \frac{\sin(\frac{1}{x})}{\frac{1}{\sqrt{x}}}$ to get it in the $\frac{\text{anything}}{\pm \infty}$ form. Executing L'Hopital's Rule I find $\lim_{x \to \infty} \sqrt{x} \sin(\frac{1}{x}) = \lim_{x \to \infty} \frac{2\cos(\frac{1}{x})}{\sqrt{x}} = \frac{0}{\infty}$ but this is yet another undetermined form is it not? It seems fairly obvious to me that the limit here is zero but also that we have an undetermined form. Do I have to keep doing rounds of L'Hopital's Rule?

• In order to use l'hopital's rule you need a form of either $\frac{0}{0}$ or $\frac{\infty}{\infty}$. And $\frac{0}{\infty}$ is not an indeterminate form. Some small number divided by a very large one is close to zero... – TravisJ May 9 '15 at 2:45

$L = \displaystyle \lim_{x \to \infty} \dfrac{1}{\sqrt{x}}\cdot \displaystyle \lim_{x \to \infty} \dfrac{\sin(\frac{1}{x})}{\frac{1}{x}} = 0\cdot 1 = 0$

No, it's not an undetermined form, because if we take absolute values and note that $0 \leq |\cos(\text{anything})| \leq 1$, then for all $x > 0$:

$\dfrac{0}{\sqrt{x}} \leq \left |\dfrac{2\cos(\frac{1}{x})}{\sqrt{x}} \right | \leq \dfrac{2}{\sqrt{x}}$.

Now, when we take $x \to \infty$, both the RHS and LHS converge to $0$, so by the squeeze theorem, the middle term converges to $0$ as well.

So doing L'Hospital's Rule, I get this: $$\lim_{x\to\infty} \frac{\cos(x^{-1}) (-x^{-2})}{\frac{-1}{2} x^{-3/2}} = \lim_{x\to\infty} \frac{2\cos \frac{1}{x}}{\sqrt{x}} = \frac{-2}{\infty} = 0$$

• @SalmonKiler I took the liberty of correcting a couple of typos. I hope that you don't mind. – Mark Viola May 9 '15 at 2:51
• @Dr.MV Not at all. – SalmonKiller May 9 '15 at 3:04

use the fact that for $x$ large $$\sin \left(\frac1x\right) = \frac1x + \mathcal O\left(\frac1{x^2}\right)$$ therefore $$\sqrt x \sin \left(\frac1x\right) = \frac1{\sqrt x} + \mathcal O\left(\frac1{x^{1.5}}\right) \to 0 \text{ as } x \to \infty.$$

• You must have some haters out there (I saw a previous post of yours that had mysterious downvotes). Apparently, people don't like series? – TravisJ May 9 '15 at 2:47
• @travisj Just yesterday I tried to address this issue of downvoting without commenting your suggestions for content improvement on META, and I was shot down. Lots of people like this status quo. I think it should be highly suggested by the site that users leave a comment on any content they downvote to help the OP improve content. For example, why the heck was this downvoted? We will never know because the idiots that downvoted didn't care to leave any comments. – layman May 9 '15 at 2:49
• It's stuff like this (the careless downvoting without trying to improve the content by commenting) that makes this website seem far from the professional mathematics site that we want it to be. – layman May 9 '15 at 2:50
• @user46944, I tried to ask the same thing once on meta and was instantly downvoted. The idea is if you force people to leave a comment, they will just make comments like "this is my comment" and still downvote (not helpful). It is sad... – TravisJ May 9 '15 at 2:55
• @user46944 I completely agree. It appears that several of us are mystified and aligned in our thinking. If one posts an answer in a sincere attempt to help the OP, then gets a down vote, what behavior does that reinforce? – Mark Viola May 9 '15 at 2:56