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 $6 L+1=2$

 

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