If $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x} \over {3{x^3}}}$$ is equal to L, then the value of (6L + 1) is
Solution
$$L = \mathop {\lim }\limits_{x \to 0} {{\left( {x + {{{x^3}} \over 6} + .....} \right) - \left( {x - {{{x^3}} \over 3}.....} \right)} \over {3{x^3}}}$$<br><br>$L = {1 \over 3}\left( {{1 \over 6} + {1 \over 3}} \right) = {1 \over 6}$<br><br>$\therefore$ $6L + 1 = 6.{1 \over 6} + 1 = 2$
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
This question is part of PrepWiser's free JEE Main question bank. 162 more solved questions on Limits, Continuity and Differentiability are available — start with the harder ones if your accuracy is >70%.