Epsilon-Delta Proof of Continuity for $\sin(x^2)$

Question

I have the function $f(x)=\sin(x^2)$, and I'm trying to prove that it is continuous on $(1,2)$. I figured using the epsilon-delta definition would be the way to go. So I have:

$\epsilon > 0$, and if $|x-a|<\delta$, then $|\sin(x^2)-\sin(a^2)|<\epsilon$

I think I should use the triangle inequality in there somewhere, but I'm not sure how.

Thanks for your time

Answer 1

It would probably be difficult to solve this directly from the $\epsilon - \delta$ definition. A better strategy would be to use the following theorem:

let $g$ be a function continuous at $x_0$ and $f$ be a function continuous at $g(x_0)$. then their composition $f \circ g$ is continuous at $x_0$.

Answer 2

Hint:

$$\sin x^2 - \sin y^2 = 2 \sin \frac{x^2 - y^2}{2}\cos \frac{x^2 + y^2}{2},$$ and $|\sin z| \leqslant |z|$