calculus - Evaluating the limit of a function using squeeze theorem

I have some function.

$$\lim_{t\to3} \ \sin{\left(\frac{1}{t-3}\right)} \ e^t \ (t-3)^2$$

I want to evaluate it using the Squeeze theorem but I do not know what it means to do this. I know that my limit is $0$ and I know that the squeeze theorem says if $f(x) \le g(x) \le h(x)$ then the limit of $g(x)$ must be the same as that of $f(x)$ and $h(x)$. However I do not understand how to implement this.

I would like to know how this works and what is happening with steps.

Answer

Hint: $-1 \le \sin{\left(\frac{1}{t-3}\right)} \le +1$