Question

$\underset{x\to 0}{lim}\frac{5(1-\cos x+4{\sin }^{-1}x-{\sin }^{2}x)}{(10{\tan }^{-1}x-x^2+x^4)}$ equals

• A. $0$
• B. $\frac{1}{2}$
• C. $2$
• D. $\frac{2}{5}$

Answer 1

The correct option is C $2$

$L=\underset{x\to 0}{lim}\frac{5(1-\cos x+4{\sin }^{-1}x-{\sin }^{2}x)}{(10{\tan }^{-1}x-x^2+x^4)}$
Dividing numerator and denominator by $x,$ we get
$L=\underset{x\to 0}{lim}\frac{5(\frac{1-\cos x}{x}+\frac{4{\sin }^{-1}x}{x}-\frac{{\sin }^{2}x}{x})}{(\frac{10{\tan }^{-1}x}{x}-x+x^3)}=\frac{5(0+4-0)}{(10-0+0)}=\frac{20}{10}=2$