Question

$\underset{h\to 0}{lim}\frac{{(a+h)}^2\sin (a+h)-a^2\sin a}{h}$ is $a(a\cos a+b\sin a).$

Find the value of $b$

Answer 1

$\underset{h\to 0}{lim}\frac{{(a+h)}^2\sin (a+h)-a^2\sin a}{h}$
$=\underset{h\to 0}{lim}\frac{{(a+h)}^2[\sin (a+h)-\sin a]+\sin a[{(a+h)}^2-a^2]}{h}$
$=\underset{h\to 0}{lim}\frac{1}{h}{(a+h)}^{2}2\cos (a+\frac{h}{2})\sin \frac{h}{2}+\underset{h\to 0}{lim}\frac{1}{h}\left(\sin a\right)[a^2+2ah+h^2-a^2]$
$=\underset{h\to 0}{lim}{(a+h)}^2\cos (a+\frac{h}{2}).\frac{\sin \left(h/2\right)}{\left(h/2\right)}+\underset{h\to 0}{lim}\left(\sin a\right)(2a+h)$
$=a^2\cos a.1+\left(\sin a\right)\left(2a\right)$
$=a(a\cos a+2\sin a).$
Using the L Hospital rule, we have
$\underset{h\to 0}{lim}\frac{{(a+h)}^2\sin (a+h)-a^2\sin a}{h}$
$=\underset{h\to 0}{lim}\frac{(a+h{)}^{2}cos(a+h)+2(a+h\left)sin\right(a+h)}{1}$
$=\left(a{)}^{2}cos\right(a+h)+2(a\left)sin\right(a)$
Hence $b=2$