Question

Evaluate the following limits:

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

Answer 1

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