calculus - what about $limlimits_{xto0}-frac{sin x}x=$?

we all know that:

$\lim\limits_{x\to0}\frac{\sin x}x=1$

so what is the negative

$\lim\limits_{x\to0}-\frac{\sin x}x=$?

i am trying to prove what about $\lim\limits_{x\to0}\frac{x^2\sin \frac 1x}{\sin^2 x}$

i got to $\frac{-x^2}{\sin^2 x} \le $ $\frac{x^2\sin \frac 1x}{\sin^2 x}\le$$\frac{x^2}{\sin^2 x}$

and $\lim\limits_{x\to0}\frac{x^2}{\sin^2 x}= \lim\limits_{x\to0}\frac{\sin x}x\lim\limits_{x\to0}\frac{\sin x}x= 1\times1=1 $

and$\lim\limits_{x\to0}-\frac{x^2}{\sin^2 x}= \lim\limits_{x\to0}\frac{\sin x}x\lim\limits_{x\to0}-\frac{\sin x}x=?$

Answer

Simply put,

$$\lim_{x\to a} (Af(x))=A\lim_{x\to a} f(x),\text{ where $A$ is a constant.}$$
So let $A=-1$ in your expression.

Edit: Let me generalize:

If $\lim_{x\to a}f(x),\lim_{x\to a}g(x)$ exist, then $\lim_{x\to a}(f(x)g(x))$ exists. So, let $g(x)=\dfrac{\sin x}{x},f(x)=-1$.