Question

${lim}_{h\to 0}\frac{\sin (a+3h)-3\sin (a+2h)+3\sin (a+h)-\sin a}{h^3}$ is equal to

• A. $\cos a$
• B. $-\cos a$
• C. $\sin a$
• D. $\sin a\cos a$

Answer 1

Answer: (B)

$-\cos a$

${lim}_{h\to 0}\frac{\sin (a+3h)-3\sin (a+2h)+3\sin (a+h)-\sin a}{h^3}$

Using LHR repeatedly.

$\Rightarrow \frac{3\cos (a+3h)-6\cos (a+2h)+3\cos (a+h)}{3h^2}\Rightarrow \frac{\cos (a+3h)-2\cos (a+2h)+\cos (a+h)}{h^2}\Rightarrow \frac{-3\sin (a+3h)+4\sin (a+2h)-\sin (a+h)}{2h}\Rightarrow \frac{-9\cos (a+3h)+8\cos (a+2h)-\cos (a+h)}{2}\Rightarrow \frac{-9\cos a+8\cos a-\cos a}{2}\Rightarrow -\cos a$