Question

$\underset{x\to 0}{lim}\frac{\sqrt{x-\sin x}}{x+{\sin }^{2}x}$ is equal to

• A. $1$
• B. $0$
• C. $\infty$
• D. none of these

Answer 1

Answer: (B)

$0$

$L={lim}_{x\to 0}\sqrt{\frac{x-\sin x}{{(x+{\sin }^{2}x)}^2}}$

$={lim}_{x\to 0}\sqrt{\frac{x-\sin x}{{x^2(\frac{{\sin }^{2}x}{x}+1)}^2}}$

$={lim}_{x\to 0}\sqrt{\frac{x-\sin x}{x^2}}\times \sqrt{\frac{1}{{(x\frac{{\sin }^{2}x}{x}+1)}^2}}$

By product rule

$L={lim}_{x\to 0}\sqrt{\frac{x-\sin x}{x^2}}\times \sqrt{\frac{1}{{(\frac{{\sin }^{2}x}{x}+1)}^2}}$

Calculating each limit seperately

$1.{lim}_{x\to 0}\sqrt{\frac{x-\sin x}{x^2}}\frac{0}{0}$ form

Using L-Hospital's rule

$=\sqrt{{lim}_{x\to 0}\left(\frac{1-\cos x}{2x}\right)}\frac{0}{0}$ form

$=\sqrt{{lim}_{x\to 0}\frac{\sin x}{2}}=\sqrt{\frac{0}{2}}=0$

$2.{lim}_{x\to 0}\sqrt{\frac{1}{{(\frac{{\sin }^{2}x}{x}+1)}^2}}$

$={lim}_{x\to 0}\frac{1}{\frac{{\sin }^{2}x}{x}+1}$

as $x\to 0,\frac{\sin x}{x}=1$

$=={lim}_{x\to 0}\frac{1}{\left(\sin x\right).1+1}=\frac{1}{0+1}=1$

$L=0\times 1=0$