0
$\begingroup$

Given the function $ f(n) = \begin{cases} \sin (1/x), & x \neq 0 \\ 0, & x=0 \end{cases} $

describe the interval on which the function is continuous.

I know that the function is continuous at all real values except $x=0$, because $y=1/x$ is continuous at all points except for $x=0$, and $\sin (1/x)$ is a composite function.

But what about the bottom half of the function? It states that there exists a point at $(0,0)$, so why wouldn't it be continuous there as well?

$\endgroup$
  • 2
    $\begingroup$ Because $\nexists \lim\limits_{x \to 0} \sin(1/x))$, so $\lim\limits_{x \to 0} f(x) \neq f(0)$ $\endgroup$ – Botond Jun 10 '18 at 19:52
  • 2
    $\begingroup$ Recall the definition of continuity requires the limit at the point be equal to the value at that point. You need to check if $$\lim_{x\to 0} \sin \frac{1}{x} = 0$$ $\endgroup$ – Osama Ghani Jun 10 '18 at 19:52
  • $\begingroup$ @Aniket Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Aug 4 '18 at 20:57
0
$\begingroup$

Note that $\lim_{x\to 0} \sin \frac{1}{x} $ doesn't exist, indeed for

  • $x_n=\frac1{2\pi n}\to 0\implies \sin\frac1{x_n}=\sin 2\pi n=0$

  • $x_n=\frac2{\pi (4n-3)}\to 0\implies \sin\frac1{x_n}=\sin \frac{\pi (4n-3)}2 =1$

then recall the definition of continuity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.