Question

Let $f\left(x\right)=\frac{1}{3}x\sin x-(1-\cos x)$. The smallest positive integer $k$ such that $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^k}\ne 0$ is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^k}=\underset{x\to 0}{lim}\frac{\frac{x\sin x}{3}-1+\cos x}{x^k}[\frac{0}{0}\text{form}]$
$=\underset{x\to 0}{lim}\frac{\frac{\sin x}{3}+\frac{x\cos x}{3}-\sin x}{k{x}^{k-1}}\left[\frac{0}{0}\text{form}\right]$
$=\underset{x\to 0}{lim}\frac{\frac{\cos x}{3}+\frac{\cos x}{3}-\frac{x\sin x}{3}-\cos x}{k(k-1)x^{k-2}}\ne 0\text{if}k=2$