Question

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

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (C)

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