Solve this following
Question:

If $\lim _{x \rightarrow 0} \frac{\sin ^{-1} x-\tan ^{-1} x}{3 x^{3}}$ is equal to $L$, then the value

of $(6 L+1)$ is

1. $\frac{1}{6}$

2. $\frac{1}{2}$

3. 6

4. 2

Correct Option: , 4

Solution:

$\lim _{x \rightarrow 0} \frac{\left(x+\frac{x^{3}}{3 !} \ldots\right)-\left(x-\frac{x^{3}}{3} \ldots\right)}{3 x^{3}}=\frac{1}{6}$

So $6 L+1=2$