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Show that the following function is not continuous at $0$.

$$f(x) = \begin{cases}0, & \text{when $x=0$} \\ \sin\left(\frac{1}{2x}\right), & \text {when $x\neq0$} \end{cases}$$

Proof.

To prove discontinuity we need to analyze the One-Sided Limits of the function.

To prove that the right limit does not exist,

let consider the sequence $\{x_n\} = \frac{1}{(2n+1)\pi}$,

and observe that $\sin (\frac{1}{2x})$ = $(-1)^n$.

Since {$x_n$} converges to 0 but $(-1)^n$ does not converge,

it follows from the Sequential Characterization of Continuity Theorem

that the right limit does not exist.

To prove that the left limit does not exist,

let consider the sequence $\{x_n\} = \frac{-1}{(2n+1)\pi}$,

and observe that $\sin \left(\frac{1}{2x}\right) = (-1)^{n+1}$.

Since $\{x_n\}$ converges to $0$ but $(-1)^{n+1}$ does not converge,

it follows from the Sequential Characterization of Continuity Theorem that the left limit does not exist.

Therefore, $f(x)$ is discontinuous at $x=0$. Q.E.D.

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    $\begingroup$ This is indeed a correct argumentation. Although actually showing that one of the one-sided limits is non zero would have been sufficient. $\endgroup$
    – b00n heT
    Oct 21, 2016 at 18:35
  • $\begingroup$ Great proof! As is mentioned in the comment above, once you show the right limit doesn't exist, you can stop there. No matter what result you get for the left limit, the left limit can't equal the right limit since the right limit doesn't exist, and hence the limit doesn't exist. $\endgroup$
    – layman
    Oct 21, 2016 at 18:43
  • $\begingroup$ perfect. nothing to add. $\endgroup$
    – hamam_Abdallah
    Oct 21, 2016 at 18:44
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    $\begingroup$ In addition to my previous comment, an equivalent characterization of continuity of a function $f$ at $x$ is that for every sequence $x_{n}$ such that $x_{n} \to x$, we get $f(x_{n}) \to f(x)$. So you could have started the proof saying "we will exhibit a sequence which converges to $x=0$, but for which the sequence of images doesn't converge to $f(0)=0$", and then your argument for the left limit would work perfectly. $\endgroup$
    – layman
    Oct 21, 2016 at 18:44
  • $\begingroup$ @b00nheT Thank you for your comment. I used both limits because there is a definition of continuity that requires only right side or left side limit, but not both. $\endgroup$
    – Beginner
    Oct 21, 2016 at 18:45

1 Answer 1

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Way overwritten. Here is a shorter proof.

There exists a sequence $x_{n}$ of positive numbers that converges to $0$ and such that the sequence $\sin(1 / x_{n}) = (-1)^n$ fails to converge. Thus, function $f(x)$ fails to be right-continuous at $0$. Since function $\sin(x)$ is odd, the sequence $-x_{n}$ analogously shows failure of left-continuity at $0$ for $f(x)$.

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    $\begingroup$ I think you're expecting too much from a beginner. OP will work their way to writing shorter proofs as they continue to learn. For their current level, their proof is fine. $\endgroup$
    – layman
    Oct 21, 2016 at 18:48
  • $\begingroup$ You are probably right. Just wanted to guard them against the Bourbakist habit of over-formalizing everything, so prevalent in modern mathematics. $\endgroup$
    – user8960
    Oct 21, 2016 at 18:48
  • $\begingroup$ @user8960 Thank you so much for your ideas. I love short proofs! I just did not see a way to do it. $\endgroup$
    – Beginner
    Oct 21, 2016 at 19:08
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    $\begingroup$ @user8960 Thank you! Very interesting! $\endgroup$
    – Beginner
    Oct 21, 2016 at 19:17
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    $\begingroup$ Here's an even shorter proof ;). Since $f(x) = 1$ and $-1$ in every neighbourhood of $0$, $\omega_f(0) = 2 \neq 0$, QED $\endgroup$
    – MathematicsStudent1122
    Oct 21, 2016 at 21:57

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