so for the the above it should be $\displaystyle cos(x^2)*2x/1$ which would be 1*0/1=0
multiplying by x/x we would get $\displaystyle xsin(x^2)/x^2$ which is equal is also equal to zero too but how? I feel like i'm missing some obvious step for version where we multiply by x/x
also is $\displaystyle sin(x^2)/x=sin^2x/x=(sinx)^2/x$ ? or is it just the last two that are the same?
$\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\frac{\sin{\left(x^2\right)}}{x} &= \lim_{x \to 0}\frac{x\sin{\left(x^2\right)}}{x^2} \\ &= \lim_{x \to 0} x \cdot \lim_{x \to 0}\frac{\sin{\left(x^2\right)}}{x^2} \\ &= 0 \cdot 1 \\ &= 0 \end{align*}$
Also, no, $\displaystyle \displaystyle \begin{align*} \sin{\left(x^2\right)} \neq \sin^2{x} \end{align*}$, rather $\displaystyle \displaystyle \begin{align*} \sin^2{x} = \left(\sin{x}\right)^2 \end{align*}$.
We have:
$\displaystyle \lim_{x\to0}\frac{x\sin(x^2)}{x^2}=\lim_{x\to0}x \cdot \lim_{x\to0}\frac{\sin(x^2)}{x^2}$
For the second limit let $\displaystyle u=x^2$ and as $\displaystyle x\to0$ we have $\displaystyle u\to0$ so we may write:
$\displaystyle \lim_{x\to0}x\cdot\lim_{u\to0}\frac{\sin(u)}{u}=0 \cdot1=0$
I am assuming you have been given $\displaystyle \lim_{u\to0}\frac{\sin(u)}{u}=1$ to use for such limits.