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



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

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


Compositions of continuous functions are continuous
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.
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:


Compositions of continuous functions are continuous
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.