I've seen a few questions posted before about mine, but this is a bit different.

The original form of the question can be found here: http://librarun.org/book/10452/159.

It says that prove by contradiction that $f(x)=\sin\left(\frac{1}{x}\right)$ has no limit value as $x$ $\rightarrow$ $0$.

I attempted to take $$x=\frac{1}{n\pi}$$ in which case $f(x)=0$ for every integer $n$. After, I write $x$ as follows $$x=\frac{1}{\frac{\pi}{2}+n\pi}$$

in which case we have $f(x)=1$ or $-1$ for every integer $n$. How do I proceed from here? Note that until this question writer did not mention about inifinite limits and limits of composite functions and sequences.


If it is not true that "for every $\epsilon > 0$ there is some $\delta > 0$ such that if $0 < |x-a| < \delta$ then $|f(x) - L| < \epsilon$", then

$$\color{blue}{\text{ there is some } \epsilon > 0 \text{ such that for every }\delta > 0 \text{ there is some } x \text{ which satisfies } \\ 0 < |x - a| < \delta \text{ but not } |f(x) - L| < \epsilon}$$

So for your limit with $f(x) = \sin(1/x)$, suppose we want to show it does not approach $0$ near $0$, consider $\epsilon = 1/2$ and note that for every $\delta > 0$ there is some $x$ with $0 < |x - 0| < \delta$ but not $|\sin(1/x) - 0| < 1/2$. Namely, $x = 1/(\pi/2 + 2\pi n )$ which satisfies $0 < |x-0| < \delta$ for some sufficiently large $n$.

More generally, any limit $L$ must lie in the interval $[-1,1]$ as that is the range of the function $\sin$. For any such $L \in [-1,1]$, there is some $x$ with $0 < |x - 0| < \delta$ but not $|\sin(1/x) - L| < 1/2$, namely

$$x = \frac{1}{\arcsin(L) \pm \pi/2 + 2\pi n}$$ for some sufficiently large $n$ and the appropriate choice of sign. (The easiest way to see this might be to look at the graphs of $y = \left|\sin(x) - \sin(x \pm \pi/2)\right|$.)

Added: Explicitly

  • For $L \in [0,1]$ choose the $-$ sign, i.e., $$x = \frac{1}{\arcsin(L) - \pi/2 + 2\pi n}$$

    Then $$|f(x) - L| = |-\cos(\arcsin(L))-L| = \sqrt{1 - L^2} + L$$ which has minimum value of $1$ on the domain $[0,1]$.

  • For $L \in [-1,0)$, chose the $+$ sign. Then $$|f(x) - L| = |\cos(\arcsin(L)) - L| = \sqrt{1-L^2} - L$$ which likewise has minimum of $1$ on $[-1,0)$.

  • $\begingroup$ why have you chosen epsilon to be $1/2$? $\endgroup$
    – frosh
    Jul 1 '15 at 15:23
  • $\begingroup$ It's necessary to show that we can find some $x$ in a neighborhood of $a$ for which the function is some $\epsilon$ distance away. A value of $\epsilon=1/2$ is a convenient choice for the rest of the argument. $\endgroup$
    – Simon S
    Jul 1 '15 at 15:27
  • $\begingroup$ I now understood the proof that states limit is not equal to zero. But still the general case seems confusing. $\endgroup$
    – frosh
    Jul 1 '15 at 15:30
  • $\begingroup$ It might help to play with this graph in WolframAlpha: bit.ly/1JvCUaG $\endgroup$
    – Simon S
    Jul 1 '15 at 15:47
  • $\begingroup$ why are we lookin' at this graph? $\endgroup$
    – frosh
    Jul 1 '15 at 16:20

Hint: for $f:D\rightarrow \mathbb R$ with $(a,x_0)\subset D, (b,x_0)\subset D$ (assuming $a<x_0<b$) the following are equivalent:

  1. $\lim\limits_{x\to x_0} f(x)=L$
  2. For every sequence $(x_n)_{n\in\mathbb N}$ in $D\setminus\{x_0\}$ with $x_n\rightarrow x_0$ it follows that $f(x_n)\rightarrow L$.
  • $\begingroup$ But, I have not familiarity with sequence concept. $\endgroup$
    – frosh
    Jul 1 '15 at 14:58
  • 1
    $\begingroup$ @MertAktaş What definition of limit do you know? This might be helpful in explaining the answer to you. $\endgroup$
    – Wojowu
    Jul 1 '15 at 15:04
  • $\begingroup$ I know the classical epsilon-delta definition. $\endgroup$
    – frosh
    Jul 1 '15 at 15:04
  • $\begingroup$ Do you know anything about sequences? Because the proof for my hint is very easy and should be a nice excercise. Or do you know anything else about limits you can use? I'd rather not use $\varepsilon-\delta$ if I can avoid it... $\endgroup$
    – Hirshy
    Jul 1 '15 at 15:11

If you set $a_n=\frac{1}{2\pi n}$ and $b_n=\frac{1}{\frac{\pi}{2}+2\pi n}$ you'll get that $a_n\to 0$, $b_n\to 0$ but $$0=\lim_{n\to\infty }f(a_n)\neq \lim_{n\to\infty }f(b_n)=1,$$ what prove the claim.

  • $\begingroup$ I wrote in advance that "Note that until this question writer did not mention about inifinite limits and limits of composite functions and sequences.". $\endgroup$
    – frosh
    Jul 1 '15 at 15:33
  • $\begingroup$ No, you have taken $x_n=\frac{\pi}{2}+n\pi$, but $f(x_n)$ doesn't converge so you don't need to take $y_n=\frac{1}{n\pi}$ because you found a sequence $(x_n)$ such that $x_n\to 0$ and $(f(x_n))$ doesn't converge. But you didn't conclude, and thus I gave you a way to justify by choosing other sequences. After, I don't understand what you say by "writer didn't mention about infinite limit of composite function and sequences", it's of course when $n\to+\infty $ for the sequences, and moreover it's obvious that $\sin(1/a_n)=\sin(2\pi n)$. $\endgroup$
    – Surb
    Jul 1 '15 at 15:40

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