For what values of $a$ and $b$ is the function $f(x)=x^a\sin(1/x^b)$ for $x \ne 0$, $f(0)=0$, continuous?

Question

Consider the function $f(x) = \begin{cases}x^a\sin({1\over x^b})&\space x \ne 0 \\ 0& \space x = 0\end{cases}$

As far as I can tell, the function is continuous everywhere so long as a and b are positive integers, but I'm not sure if this is correct or not. The only place where I see it potentially not being continuous is at 0. However, using squeeze theorem, the $\lim_{x \to 0} f(x) = \lim_{x \to 0} x^a\sin({1\over x^b}) = 0$ for both odd and even a's and b's.

So am I correct?

Answer 1

Edit: I have made my previous answer a little bit well-ordered.

Your answer is partially true. We have many cases.

1) $a=0$. You can easily check that only for $b<0$ we have continuity.

2) $a>0$. As you said the squeeze theorem gives the answer: continuous.

3) $a<0$.

• $b=0$, then $\lim_{x\to 0}f(x)$ does not exist (why?), so not continuous.
• $b>0$, then take the sequence $x_n= \frac{1}{\sqrt[b]{\frac{1}{2}\pi +2n\pi}}$. So that $x_n\to 0$ and $\sin(x_n)=1$ for all $n\in \mathbb{N}$. Furthermore $(x_n)^a = (\frac{1}{2}\pi +2n\pi)^{-\frac{a}{b}} \to \infty$, so that $f(x_n) \to \infty$ which implies: not continuous.
• $b<0$, then we have: \begin{align} \lim_{x\to 0} f(x) = \lim_{x\to 0} x^a\sin(x^{-b}) \stackrel{Taylor}{=} \lim_{x\to 0} x^a \left( x^{-b} + O(x^{-3b}\right) = \lim_{x\to 0} x^{a-b} + O(x^{a-3b}) \end{align} If $a=b$ then we have $\lim_{x\to 0} f(x)=1$, if $a>b$ then we have $\lim_{x\to 0} f(x) =0$ and if $a<b$ the limit does not exist.

Make the final conclusion yourself. I hope this helps you. Let me know if it is not clear yet.

Answer 2

The function $f$ is continuous at $x=0$ if and only if $a> b$ or $ a>0$.

In fact , using this inequality, $$|\sin x|\le |x|$$ together with $|\sin x|\le 1$ we arrive at

$$|f(x)|\le\min( |x|^{a-b},|x|^{a})\tag{I}$$

•If $a>0$ then (I) implies that $f(x)\to0 $ as x goes to 0.

•If $a \le 0$ and $a-b>0$ then (I) implies that $f(x)\to0 $ as x goes to 0.

•otherwise the continuity does not hold.

For the continuity fails for the other case if one takes into account that

$$\lim_{x\to 0}\frac{\sin x}{x} =1$$

In the case where $a=b\le0$