Question

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

• A. $a^{2}cosa$
• B. $a^{2}cosa+2asina$
• C. $a^{2}sina+cosa$
• D. $acosa+sina$

Answer 1

Answer: (B)

We have, $\underset{h\to 0}{\lim }$$\frac{\left(a+h^2\right)sin\left(a+h\right)-a^{2}sina}{h}$
$=\underset{h\to 0}{\lim }$ $\frac{\left(a^2+h^2+2ah\right)\left[sinacosh+cosasinh\right]-a^{2}sina}{h}$
$=\underset{h\to 0}{\lim }$$[\frac{a^{2}sina\left(cosh-1\right)}{h}+\frac{a^{2}cosasinh}{h}$
$+\left(h+2a\right)\left(sinacosh+cosasinh\right)]$
$=\underset{h\to 0}{\lim }$ $\left[\frac{a^{2}sina\left(-2si{n}^2\frac{h}{2}\right)}{\frac{h^2}{2}}\cdot \frac{h}{2}\right]+$ $\underset{h\to 0}{\lim }$$\frac{a^{2}cosasinh}{h}$
$+\underset{h\to 0}{\lim }(h+2a)sin(a+h)$
$=a^{2}sina\times 0+a^{2}cosa\left(1\right)+2asina$
$=a^{2}cosa+2asina$.