Question

$\underset{h\to 0}{\lim }\frac{{\left(a+h\right)}^2\sin \left(a+h\right)-a^2\sin a}{h}$

Answer 1

$\underset{h\to 0}{\lim }\left[\frac{{\left(a+h\right)}^2\sin \left(a+h\right)-a^2\sin a}{h}\right]$

$$\begin{gathered} =\underset{h\to 0}{\lim }\left[\frac{a^2\sin \left(a+h\right)+h^2\sin \left(a+h\right)+2ah\sin \left(a+h\right)-a^2\sin a}{h}\right] \\ =\underset{h\to 0}{\lim }\left[a^2\left\{\frac{\sin \left(a+h\right)-\sin \left(a\right)}{h}\right\}+\frac{h^2\sin \left(a+h\right)}{h}+\frac{2ah\sin \left(a+h\right)}{h}\right] \\ \mathrm{Div}\text{iding and multiplying the denominator by 2:} \\ \underset{h\to 0}{\lim }\left[a^2\left\{\frac{2\cos \left({\displaystyle \frac{a+h+a}{2}}\right)\sin \left({\displaystyle \frac{a+h-a}{2}}\right)}{2\times {\displaystyle \frac{h}{2}}}\right\}+h\sin \left(a+h\right)+2a\sin \left(a+h\right)\right] \\ =\underset{h\to 0}{\lim }\left[a^2\cos \left(\frac{a+h+a}{2}\right)+h\sin \left(a+h\right)+2a\sin \left(a+h\right)\right] \\ =a^2\cos a+0+2a\sin a \\ =a^2\cos a+2a\sin a \end{gathered}$$