Question

$\displaystyle \lim_{h \to 0}$ $\frac{\left(a+h^{2}\right)sin\left(a+h\right)-a^{2}\,sin\,a}{h}=$

• A. $a^2\,cos\,a$
• B. $a^{2}\,cos\,a+2a\,sin\,a$
• C. $a^{2}\,sin\,a+cos\,a$
• D. $a\,cos\,a+sin\,a$

Answer 1

The Correct Option is B

We have, $\displaystyle \lim_{h \to 0}$$\frac{\left(a+h^{2}\right)sin\left(a+h\right)-a^{2}\,sin\,a}{h}$ $=\displaystyle \lim_{h \to 0}$ $\frac{\left(a^2+h^{2}+2ah\right)\left[sin\,a\,cos\,h+cos\,a\,sin\,h\right]-a^{2}\,sin\,a}{h}$ $=\displaystyle \lim_{h \to 0}$$\bigg[\frac{a^{2}\,sin\,a\left(cos\,h-1\right)}{h}+\frac{a^{2}\,cos\,a\,sin\,h}{h}$ $+\left(h+2a\right)\left(sin\,a\,cos\,h+cos\,a\,sin\,h\right)\bigg]$ $=\displaystyle \lim_{h \to 0}$ $\left[\frac{a^{2}\,sin\,a\left(-2\,sin^{2}\, \frac{h}{2}\right)}{\frac{h^{2}}{2}}\cdot \frac{h}{2}\right]+$ $\displaystyle \lim_{h \to 0}$$\frac{a^{2}\,cos\,a\,sin\,h}{h}$ $+\displaystyle \lim_{h \to 0}(h+2a)sin(a+h)$ $=a^{2}\,sin\,a \times 0+a^{2}\,cos\,a\left(1\right)+2a\,sin\,a$ $=a^{2}\,cos\,a+2a\,sin\,a$.