lim h tends to 0 (a +h)^2sin(a+h)-a^2sina whole divided by h

This is of the form $\frac{0}{0}$. So we can use L'Hospital's Rule to solve it.

$$\begin{gathered} \underset{h\to 0}{\lim } \frac{{\left(a+h\right)}^2\sin \left(a+h\right)-a^2\sin a}{h} \\ =\underset{h\to 0}{\lim } \frac{2\left(a+h\right)\sin \left(a+h\right) + {\left(a+h\right)}^2\cos \left(a+h\right) - 0}{1} \end{gathered}$$
$$\begin{gathered} =2\left(a+0\right)\sin \left(a+0\right) + {\left(a+0\right)}^2\cos \left(a+0\right) \\ =2a\sin a + a^2\cos a \end{gathered}$$