Question

If $\underset{x\to 0}{lim}\frac{{\sin }^{-1}x-{\tan }^{-1}x}{3x^3}$ is equal to $L$, then the value of $(6L+1)$ is:

• A. $\frac{1}{2}$
• B. $2$
• C. $\frac{1}{6}$
• D. $6$

Answer 1

The correct option is B

$2$

Explanation for the correct option.

Step 1: Apply L' Hospital's rule.

By applying limit, we can see that $\frac{{\sin }^{-1}x-{\tan }^{-1}x}{3x^3}$is of $\frac{0}{0}$form. So, we will apply L' Hospital's rule, according to which

$\underset{x\to c}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to c}{lim}\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$.

So,

$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{{\sin }^{-1}x-{\tan }^{-1}x}{3x^3}& =& \underset{x\to 0}{lim}\frac{{\displaystyle \frac{d\left({\sin }^{-1}x\right)}{dx}}-{\displaystyle \frac{d\left({\tan }^{-1}x\right)}{dx}}}{{\displaystyle \frac{d\left(3x^3\right)}{dx}}}\\ & =& \underset{x\to 0}{lim}\frac{{\displaystyle \frac{1}{\sqrt{1-x^2}}-\frac{1}{1+x^2}}}{{\displaystyle 9x^2}}\\ & =& \underset{x\to 0}{lim}\frac{1+x^2-\sqrt{1-x^2}}{9x^2\times \sqrt{1-x^2}\times \left(1+x^2\right)}\\ & =& \underset{x\to 0}{lim}\frac{1+x^2-\sqrt{1-x^2}}{9x^2}\left[As\underset{x\to 0}{lim}\sqrt{1-x^2}\times \left(1+x^2\right)=1\right]\end{array}$

This is again in the form of $\frac{0}{0}$.

Step 2: Apply L' Hospital's rule again.

$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{1+x^2-\sqrt{1-x^2}}{9x^2}& =& \underset{x\to 0}{lim}\frac{0+2x-{\displaystyle \frac{1}{2\sqrt{1-x^2}}}\left(-2x\right)}{18x}\\ & =& \underset{x\to 0}{lim}\frac{2x+{\displaystyle \frac{x}{\sqrt{1-x^2}}}}{18x}\\ & =& \underset{x\to 0}{lim}\frac{2+{\displaystyle \frac{1}{\sqrt{1-x^2}}}}{18}\\ & =& \frac{2+{\displaystyle \frac{1}{\sqrt{1-0}}}}{18}\\ & =& \frac{3}{18}\\ & =& \frac{1}{6}\end{array}$

Step 3: Find the value of $(6L+1)$.

We have

$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{{\sin }^{-1}x-{\tan }^{-1}x}{3x^3}& =& L\\ \Rightarrow L& =& \frac{1}{6}\end{array}$

So,

$\begin{array}{rcl}(6L+1)& =& 6\left(\frac{1}{6}\right)+1\\ & =& 2\end{array}$

Hence, option B is correct.