JEE Mains · Maths · STD 11 - 12. limits
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
- A \(\frac{1}{6}\)
- B \(\frac{1}{2}\)
- C \(6\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0} \frac{\left(x+\frac{x^{3}}{3 !} \ldots\right)-\left(x-\frac{x^{3}}{3} \ldots\right)}{3 x^{3}}=\frac{1}{6}\) So \(6 L+1=2\)
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