Question

Solve: $x\stackrel{lim}{\to }0\left[\frac{x^2}{{\int }_0^x{\tan }^{-1}tdt}\right]=$

• A. $1$
• B. $2$
• C. $0$
• D. $4$

Answer 1

The correct option is B $2$

Given, $x\stackrel{lim}{\to }0\left[\frac{x^2}{{\int }_0^{x}ta{n}^{-1}tdt}\right]$

$=\frac{0}{0}form$

$=x\stackrel{lim}{\to }0\frac{\frac{d}{dx}\left(x^2\right)}{\frac{d}{dx}\left[{\int }_0^{x}ta{n}^{-1}tdt\right]}$

$=x\stackrel{lim}{\to }0\frac{2x}{ta{n}^{-1}x}$

$=x\stackrel{lim}{\to }0\frac{2x}{ta{n}^{-1}x}=\frac{0}{0}form$

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

$=x\stackrel{lim}{\to }0\left[2\right(1+x^2\left)\right]=2$