# Prove $\lim _{x \to 0} \sin(\frac{1}{x}) \ne 0$

Prove $$\lim _{x \to 0} \sin\left(\frac{1}{x}\right) \ne 0.$$

I am unsure of how to prove this problem. I will ask questions if I have doubt on the proof. Thank you!

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@Argon Careful with the variable and the path. You probably want $0^+$. –  Pedro Tamaroff Oct 15 '12 at 2:18
May I mention that one ought not to use the operator $\lim$ when, as is the case here, limit there is not. For this function $f$, $\lim\limits_{x\to0}f(x)\ne0$ is not true. –  Did Oct 15 '12 at 6:53
@did: I almost agree with what you say here. One may use $\lim\limits_{x\to0}f(x)$, but only to say that this limit does not exist. Where I do agree 100%: writing $\lim\limits_{x\to0}f(x)\ne0$ make no sense if the limit doesn't exist, so one shouldn't write that down. –  Hendrik Vogt Oct 15 '12 at 9:12
@Hendrik Any source for the use of $\lim\limits_{x\to0}f(x)$ when $\lim\limits_{x\to0}f(x)$ does not exist? Which seems weird, if you ask me... –  Did Oct 15 '12 at 9:58
@did: On the spur of the moment I can only think of this cinematic source ... –  Hendrik Vogt Oct 15 '12 at 10:17

HINT

Consider the sequences $$x_n = \dfrac1{2n \pi + \pi/2}$$ and $$y_n = \dfrac1{2n \pi + \pi/4}$$ and look at what happens to your function along these two sequences. Note that both sequences tend to $0$ as $n \to \infty$.

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We don't need to consider both sequences (since we're only trying to prove it doesn't converge to $0$, not that it doesn't converge at all). Either one will do the trick. –  Cameron Buie Oct 15 '12 at 1:53
@CameronBuie True. If I had written just the first sequence, I thought the OP's next question would be if the limit of the function is $1$. Hence, I considered both the sequence to show that the limit doesn't exist. –  user17762 Oct 15 '12 at 4:48
Fair point. I ought to have assumed there was a method to your (apparent) madness. ;) –  Cameron Buie Oct 15 '12 at 6:50

in order to prove the limit $\not=0$, only need to find a sequent $\{x_n\}\rightarrow0$,but $\sin(x_n) \not\rightarrow 0$. and the below is the construction.

$$x_n=\frac{1}{\frac{\pi}{2}+2n\pi}$$

$$\sin\left(\frac{1}{x_n}\right)=1$$

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I think the downvote is there because you didn't add "words" to your answer. You've written equations but you don't say why you've written them. If you want your answers to be upvoted and accepted by the population of math.stackexchange, put some work into it. =) –  Patrick Da Silva Oct 15 '12 at 1:53
OK,thanks for your advice.:) –  Laura Oct 15 '12 at 3:48
Yes, that is indeed a little bit more appropriate. Sad part is that at this point there are already many answers up above. I upvoted you for the effort though. –  Patrick Da Silva Oct 15 '12 at 4:01

To show that, some limit of some function does not exist it is better often use Heine definition of limit, which states that a function $f(x)$ has a limit $L$ at $x = a$, if for every sequence $\{x_n\}$ , which has a limit at $a$, the sequence $f(\{x_n\})$ has a limit $L$.

Thus, by the above definition, you need to find(it is already done for you, in above answer) sequences $\{x_n\}$ and $\{y_n\}$ such that $\lim x_n=\lim y_n=0$, but $f(x_n)\neq f(y_n)$.

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Here's an intuitive answer, just to shake things up. The limit of a function as it approaches some point is the value you expect the function to take at that point$^1$ based on what it does nearby the point. If you can't form any expectations about the function's value, the limit doesn't exist.

Now take a look at the function in question:

Can anyone possibly form expectations about where that function should be at $0$?

Of course, once you have the intuition, you have to actually write down the proof. For that, you need to remember the definition of the limit and formalize why you can't figure out where $\sin(x^{-1})$ "should" be at $0$.

1. (Be careful, though! The function might not even be defined at the point! We're just forming an expectation of what we think the function should be, if it were to exist. We don't know anything about what the function actually does.)
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Your intuitive idea is just wrong. The limit of a function as it approaches some point is not the value you expect the function to take at that point. It's the value you expect the function to approach when you get closer and closer to that point. Your idea is correct for continuous functions, but not all functions are continuous and the concept of limit is precisely introduced to study the continuity of functions. So it is a very bad idea to give the intuition this way. –  Patrick Da Silva Oct 15 '12 at 1:52
I disagree. Since limit of $f$ at $x_0$, if it exists, is the value that $f(x)$ approaches as $x$ approaches $x_0$, what else would one expect $f(x_0)$ to be? I don't understand your objection to using this intuition to introduce continuity. Intuitively, a function is continuous at $x_0$ if it adopts the value it "should" adopt at $x_0$ based on its nearby behavior. It's discontinuous if it violates expectations, i.e., if $f(x_0) \neq \lim_{x\to x_0}f(x)$. –  Neal Oct 15 '12 at 2:00
My objection is in your comment ; you add "if it exists". What if it doesn't? Then your intuition is wrong. To introduce continuity it's okay to describe it as is, but in the current context we speak of non-continuous functions, so it's not really relevant. And note that we are also working with an undefined function at $0$ : how can you even speak of $f(x_0)$ in this case? Speaking of limit without speaking of "getting closer to" is not a good idea. –  Patrick Da Silva Oct 15 '12 at 2:06
@Neal To put it simple, what is $\lim_{x \to 0} \frac{\sin(x)}{x}$? Is it $\sin(0)/0$? –  N. S. Oct 15 '12 at 2:08
My only objection now is that you speak of "the function's value" even if it does not need to have one at $x_0$. Everything in the concept of limit is about approaching a point at which the function does not need to be defined. That is why I don't like your intuitive idea. Sorry if I'm being arrogant ; I only mean to explain my point. –  Patrick Da Silva Oct 15 '12 at 2:39

As a complement to Marvis' answer and to Neal's drawing, consider $$x_n(\delta) = \frac 1{2n\pi + \delta}.$$ Then $$f(x_n) = \sin(2 n \pi + \delta) = \sin (\delta) \underset{n \to \infty}{\longrightarrow} \sin(\delta).$$ Choosing $\delta$ such that $\sin(\delta) = y \in [-1,1]$, we see that using an appropriate sequence $x_n(\delta)$ converging to $0$, we can approach any value between $[-1,1]$. Since $|f(x_n)| \le -1$, this explains the ugly behavior of the function near $0$ as seen in Neal's graph.

Hope that helps,

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