The question is as follows,

Determine the values of $n$ for which the function, $$f(x) = \begin{cases}x^n\sin\left(\frac1x\right) & ,x\neq 0 \\ 0 & ,x=0\end{cases}$$ is continuous at $x=0$

The way I tried to solve it is by using inequalities, starting with, $$-1 \leq \sin \left(\frac1x\right) \leq 1$$ $$-x^n \leq x^n\sin \left(\frac1x\right) \leq x^n$$ $$\lim_{x\rightarrow 0} -x^n \leq \lim _{x\rightarrow0} f(x) \leq \lim_{x\rightarrow 0} x^n $$ Which gives us, continuity $\forall \;n$
This seems right but I'm not sure, could someone confirm it and also is there a better method?

  • 2
    $\begingroup$ Well done! Your answer is correct. The function is continuous for all $n \in \Bbb N$ $\endgroup$ Commented Jun 15, 2020 at 17:28
  • 3
    $\begingroup$ what if $n=0{}$? $\endgroup$ Commented Jun 15, 2020 at 17:28
  • $\begingroup$ An interesting extension is to answer the same question, but instead of asking about continuity, consider whether the 1st derivative exists. What about the 2nd derivative? n-th derivative? $\endgroup$ Commented Jun 15, 2020 at 17:29
  • $\begingroup$ @BenjaminWang I will try that too :) $\endgroup$ Commented Jun 15, 2020 at 17:32
  • $\begingroup$ @AnginaSeng I'm not sure since that would be an indeterminant form right? ($0^0$) $\endgroup$ Commented Jun 15, 2020 at 17:33

2 Answers 2


Let $f_n(x) = x^n\,\sin(1/x)$. About the easy cases first, this function is continuous when $x$ is away from $0$ by as a composition of continuous functions, and as you say, when $n>0$, $|f_n(x)| ≤ |x|^n \underset{x\to 0}{\to} 0$, so the function is also continuous in $0$.

Now if $n=0$. Take the sequence $$ x_k = \frac{1}{2πk+\pi/2} $$ Then $x_k\underset{k\to \infty}{\to} 0$, but $$ f(x_k) = \sin(2πk+\pi/2) = \sin(\pi/2) = 1 $$ does not converge to $0$, so the function is discontinuous in $0$ by the sequential definition of the limit (the same is true by the way if $n<0$).

  • $\begingroup$ Where do you see any $(x_k)^n$? The only case when I take $x_k$ is when $n=0$, so $x^n=1$ for any $x>0$ (so if you prefer, the sequence $(x_k)^0 =1$ so it converges to $1$, was that your problem?) $\endgroup$
    – LL 3.14
    Commented Jun 16, 2020 at 16:21
  • $\begingroup$ Yes, I deleted my comment, I came to the same conclusion. Although I am wondering now, how did you come with the expression for $x_k$? After all we could take $x_k$ as $\frac{1}{2\pi k + \pi/6}$ $\endgroup$ Commented Jun 16, 2020 at 19:54
  • $\begingroup$ Yes, of course, you can take a lot of different sequences that will converges to whatever value between $-1$ and $1$. I just chose $1$ to take a simple case, since you just need to find one sequence along which the function is not converging to $0$. $\endgroup$
    – LL 3.14
    Commented Jun 16, 2020 at 20:54

Define $f_n(x)=x^n\sin\frac{1}{x}$, $x\neq0$.

For any $n>0$, setting $f_n(0)=0$, one obtains a continuous function since $|f_n(x)|\leq x^n\xrightarrow{x\rightarrow0}0$.

If $n\leq0$, the discontinuity at $x=0$ is not removable as can can see by choosing sequences $x_n=\frac{2}{\pi(2n+1)}$ and $y_n=\frac{1}{n\pi}$. Both sequences converge to $0$ but $f(y_n)=0$ while $f(y_n)$ is highly oscillatory.


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