Question

$\underset{x\to 0}{lim}\frac{\sin x-x+\frac{x^3}{6}}{x^5}=$

• A. $\frac{1}{140}$
• B. $\frac{1}{5}$
• C. $\frac{1}{60}$
• D. $\frac{1}{120}$

Answer 1

The correct option is D $\frac{1}{120}$

$\underset{x\to 0}{lim}\frac{\sin x-x+\frac{x^3}{6}}{x^5}$

It is of the form $\frac{0}{0}$, so applying L-Hospital's rule

$=\underset{x\to 0}{lim}\frac{\cos x-1+\frac{x^2}{2}}{5x^4}$

Again, applying L-Hospital's rule

$=\underset{x\to 0}{lim}\frac{-\sin x+x}{20x^3}$

Again, applying L-Hospital's rule

$=\underset{x\to 0}{lim}\frac{-\cos x+1}{60x^2}$

Again, applying L-Hospital's rule

$=\underset{x\to 0}{lim}\frac{\sin x}{120x}=\frac{1}{120}$