Prove continuity by the epsilon-delta definition

Question

Let $a>0$, $b>0$ and $f:\mathbb{R} \to \mathbb{R}$ define by \begin{align} f(x) = \left\{ \begin{array}{ll} x^a \sin{x^{-b}} \quad &\text{if } x \ne 0, \\ 0 \quad &\text{if } x = 0 \end{array} \right. \end{align} I have to find all values of $a$ and $b$ when $f$ is continuous.

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We know that $f$ is continues on $(\infty, 0)$ if $\forall x<0$, $\forall \epsilon > 0$, $\exists \delta > 0$ and $\forall y<0$ such as $$0 < |x-y| < \delta \implies |x^a\sin{x^{-b}} - y^a\sin{y^{-b}}| < \epsilon$$

But I don't know how to approach the right side of the implication to isolate $|x-y|$. Can you help me?

Answer 1

The question is unclear, but it sounds like you want to show that the function is continuous on $(-\infty, 0)$, which does not require epsilons and deltas. Just use the facts that:

1. Compositions of continuous functions are continuous
2. Quotients of continuous functions are continuous if the divisor doesn't attain $0$ on the interval.

Then $\dfrac{1}{x^b}$, $x^a$, and $\sin{x}$ are continuous on $(-\infty, 0)$, so $x^a \sin{\dfrac{1}{x^b}}$ is also continuous.