4
$\begingroup$

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!

  • 1
    @Argon Careful with the variable and the path. You probably want $0^+$.2012-10-15
  • 0
    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**.2012-10-15
  • 0
    @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.2012-10-15
  • 0
    @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...2012-10-15
  • 0
    @did: On the spur of the moment I can only think of [this cinematic source](http://www.youtube.com/watch?v=oDAKKQuBtDo) ...2012-10-15
  • 0
    @Hendrik Veeeeery convincing, I must admit...2012-10-15

5 Answers 5

5

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 $n \to \infty$.

  • 1
    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.2012-10-15
  • 0
    @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.2012-10-15
  • 0
    Fair point. I ought to have assumed there was a method to your (apparent) madness. ;)2012-10-15
5

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:

Topologist's sine curve!

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.)
  • 1
    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.2012-10-15
  • 4
    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 $f(x_0) \neq \lim_{x\to x_0}f(x)$.2012-10-15
  • 1
    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.2012-10-15
  • 0
    @Neal To put it simple, what is $\lim_{x \to 0} \frac{\sin(x)}{x}$? Is it $\sin(0)/0$?2012-10-15
  • 0
    @N.S. The limit is $1$. Of course it's not $\sin(0)/0$. Maybe I'm just dense today, but I don't see your point.2012-10-15
  • 0
    @PatrickDaSilva Why is the intuition wrong if the limit doesn't exist? That would mean that it's not possible to form any expectations about the function's value based on what it does nearby. This does not require assuming that $f$ is even defined at $x_0$.2012-10-15
  • 0
    (I added "based on what it does nearby the point" to the answer to convey that the limit is determined by examining nearby values of the function.)2012-10-15
  • 0
    @Neal : Yes, that would be a little better. Now the only reason I would not use your approach is because I don't expect functions to be continuous in general, but at this point it's only a matter of taste, so I don't want to argue. Thanks for caring about my comment though, I appreciate it.2012-10-15
  • 1
    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.2012-10-15
  • 0
    Okay, I think I see what you're saying. Let me see if I can edit the post to address your point.2012-10-15
3

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.

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

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

  • 0
    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. =)2012-10-15
  • 0
    OK,thanks for your advice.:)2012-10-15
  • 0
    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.2012-10-15
1

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 $f(x_n)\neq f(y_n)$.

1

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,