Question

Evaluate $\underset{n\to \infty }{lim}\frac{\sin x-x+\frac{x}{6}}{x^5}$ using series expansion

• A. $\frac{3}{120}$
• B. $\frac{1}{120}$
• C. $\frac{1}{130}$
• D. $\frac{1}{2400}$

Answer 1

Answer: (B)

$\frac{1}{120}$

$\underset{n\to \infty }{lim}\frac{\sin x-x+\frac{x}{6}}{x^5}$

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$

$=\underset{n\to \infty }{lim}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-x+\frac{x}{6}}{x^5}$

$=\underset{n\to \infty }{lim}(\frac{1}{6x^4}-\frac{1}{3!x^2}+\frac{1}{5!})$

$\underset{n\to \infty }{lim}\frac{1}{x}=0$

$=\underset{n\to \infty }{lim}(0-0+\frac{1}{5!})$

$=120$