Question

$\underset{x\to 0}{lim}\frac{1-{\cos }^{2}x}{x\sin x\cos x}$ is equal to .

• A. $1$
• B. $\frac{3}{5}$
• C. $\frac{3}{2}$
• D. $\frac{5}{4}$

Answer 1

The correct option is A $1$

${lim}_{x\to 0}\frac{1-\cos {}^{2}x}{x\sin x\cos x}$

$\Rightarrow {lim}_{x\to 0}\frac{\sin {}^{2}x}{x\sin x\cos x}$

$\Rightarrow {lim}_{x\to 0}\frac{\sin x}{x\cos x}\left(\frac{0}{0}form\right)$

Using L-Hopital rule.

$\Rightarrow {lim}_{x\to 0}\frac{\cos x}{\cos x-x\sin x}$

$\Rightarrow \frac{1}{1-0}=1$