$x^2\sin(x)$ continuous using $\epsilon\delta$ method I would like to prove that the function : $f: \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = x^2 \sin(x)$ is continuous.
I know that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous iff : 
$\forall x \in \mathbb{R}, \forall \epsilon > 0, \exists \beta > 0, \forall y \in \mathbb{R}, \mid x-y \mid \leq \beta\Rightarrow \mid f(x)-f(y)| < \epsilon$
That's why here in order to proove that $f$ is continuous I must find a $\beta$ such that : 
$\mid x - y \mid \leq \beta \Rightarrow \mid x^2\sin(x) -y^2\sin(y) \mid < \epsilon$
I tried to use the fact that $\sin(x) \in [-1, 1]$ but it doesn't seem to help a lot.

Hint use $$| x^2\sin x -y^2\sin y | =|x^2\sin x-y^2\sin x+y^2\sin x-y^2\sin y|$$
$$\leq |x^2\sin x-y^2\sin x|+|y^2\sin x-y^2\sin y|=|x^2-y^2||\sin x|+y^2|\sin x-\sin y|$$
$$=|x-y||x+y||\sin x|+2y^2 |\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)|$$
$$=|x-y||x+y||\sin x|+2y^2 |\sin\left(\frac{x-y}{2}\right)||\cos\left(\frac{x+y}{2}\right)|$$
Now, use the upper bound for sine and cosine
$$| x^2\sin x -y^2\sin y |\leq |x-y||x+y|+2y^2 |\sin\left(\frac{x-y}{2}\right)|.$$
Can you complete it from here?
