Q. $\lim_{h \rightarrow 0}\frac{(a + h)^{2}\sin{(a + h) - a^{2}\sin a}}{h}$ is equal to

A) $2a\sin a$
B) $a^{2}\cos a$
C) $a^{2}\cos{a + 2a\sin a}$
D) None of these
Explanation

(c)
Here, $\lim_{h \rightarrow 0}\frac{(a + h)^{2}\sin{(a + h) - a^{2}\sin a}}{h}$
$= \lim_{h \rightarrow 0}\left\lbrack \frac{a^{2}\left\{ \sin{(a + h) - \sin a} \right\}}{h} + \frac{h\{ 2a\sin{(a + h) + h\sin{(a + h)\}}}}{h} \right\rbrack$
$= \lim_{h \rightarrow 0}\frac{a^{2}.2\cos{\left\lbrack a + \frac{h}{2} \right\rbrack.\sin\frac{h}{2}}}{2.\frac{h}{2}} + \lim_{h \rightarrow 0}{(2a + h)\sin{(a + h)}}$
$= a^{2}\cos{a + 2a\sin a}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.