It doesn't work that way.
You have to use partial fractions to split the limit into a sum, not a product.
$\dfrac{\sin(5x)}{2x(x-2)}= \dfrac{\sin (5 x)}{4 (x-2)}-\dfrac{\sin (5 x)}{4 x}$
$\begin{align*}
&\lim \limits_{x\to 0}\dfrac{\sin(5x)}{2x(x-2)} = \\ \\
&\lim \limits_{x\to 0}\left(\dfrac{\sin (5 x)}{4 (x-2)}-\dfrac{\sin (5 x)}{4 x}\right) = \\ \\
&\lim \limits_{x\to 0} \dfrac{\sin (5 x)}{4 (x-2)} - \lim \limits_{x\to 0}\dfrac{\sin (5 x)}{4 x} = \\ \\
&\dfrac{0}{-8} - \lim \limits_{x\to 0} \dfrac 5 4 \dfrac{\sin(5x)}{5x} = \\ \\
&0 - \dfrac 5 4 ( 1) = \\ \\
&-\dfrac 5 4
\end{align*}$
But how come what I did does not work. And its not the multiplication that's wrong, she said that is fine. It is the fact that I split the limit and solved one part of it before the other and I don't understand why that is wrong.
No you don't, the limit of a product is equal to the product of the limits.
The point is that the two functions in the product are approaching 0 at the same time, so they should be evaluated at the same time. Computationally it doesn't make any difference, but conceptually it does.