Question

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

• A. 1/6
• B. 6
• C. 2
• D. 1/2

Hint:

Alternatively, apply L'Hôpital's rule thrice, but Taylor expansions are much faster for $x \to 0$ limits involving inverse trig functions.

Answer 1

The Correct Option is C

Step 1: Use expansions for small $x$: $\sin^{-1}x = x + \frac{x^3}{6} + \frac{3x^5}{40} + .......$ $\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - .......$
Step 2: Substitute into the limit: $L = \lim_{x \to 0} \frac{(x + \frac{x^3}{6}) - (x - \frac{x^3}{3})}{3x^3} = \lim_{x \to 0} \frac{\frac{x^3}{6} + \frac{x^3}{3}}{3x^3}$
Step 3: $L = \frac{1/6 + 1/3}{3} = \frac{1/2}{3} = 1/6$.
Step 4: Calculate $6L + 1 = 6(1/6) + 1 = 1 + 1 = 2$.