Question

Solve $\underset{x\to 0}{lim}\frac{{\sin }^{-1}x-{\tan }^{-1}x}{x^3}$ equal

• A. $1$
• B. $-1$
• C. $1/2$
• D. $-3/2$

Answer 1

Answer: (C)

$1/2$

$\underset{x\to 0}{lim}\frac{{\sin }^{-1}x-{\tan }^{-1}x}{x^3}$

Applying L'Hopital here,limit becomes
$\underset{x\to 0}{lim}\frac{\frac{1}{1-\sqrt{x^2}}-\frac{1}{1+x^2}}{3x^2}$

$=\underset{x\to 0}{lim}\frac{(1+x^2)-\sqrt{1-x^2}}{3x^2(1+x^2)\sqrt{1-x^2}}$

Multiply numerator and denominator by $(1+x^2)+\sqrt{1-x^2}$ and we get
$=\underset{x\to 0}{lim}\frac{(1+x^2{)}^2-(1-x^2)}{3x^2\left(\right(1+x^2)+\sqrt{1-x^2})(1+x^2)\sqrt{1-x^2}}$

$=\underset{x\to 0}{lim}\frac{3+x^2}{3\left(\right(1+x^2)+\sqrt{1-x^2})(1+x^2)\sqrt{1-x^2}}$

Putting in $x=0$ into the above equation, We get $\frac{1}{2}$.