Question

Evaluate the limit: \[ \lim_{x \to 1} \frac{\sqrt{x} - 1}{(\cos^{-1} x)^2} =\] 

• A. \( -\frac{1}{4} \)
• B. \( \frac{1}{2} \)
• C. \( -\frac{1}{2} \)
• D. \( \frac{1}{4} \)

Hint:

For limits involving inverse trigonometric functions, use the first-order approximations \( \cos^{-1} x \approx \frac{\pi}{2} - \sqrt{2(x - 1)} \) and square expansions for simplification.

Answer 1

The Correct Option is A

Step 1: Apply First-Order Approximations 
Using the first-order Taylor series expansion near \( x = 1 \): \[ \sqrt{x} = 1 + \frac{(x - 1)}{2} + O((x - 1)^2) \] \[ \cos^{-1} x = \frac{\pi}{2} - \sqrt{2(x - 1)} + O((x - 1)^{3/2}). \] 
Step 2: Simplify the Numerator and Denominator 
The numerator: \[ \sqrt{x} - 1 = \frac{(x - 1)}{2} + O((x - 1)^2). \] The denominator: \[ (\cos^{-1} x)^2 = \left(\frac{\pi}{2} - \sqrt{2(x - 1)} + O((x - 1)^{3/2})\right)^2. \] Expanding the square: \[ (\cos^{-1} x)^2 = \frac{\pi^2}{4} - \pi\sqrt{2(x - 1)} + 2(x - 1) + O((x - 1)^{3/2}). \] 
Step 3: Compute the Limit 
Dividing the numerator by the denominator: \[ \lim_{x \to 1} \frac{\frac{(x - 1)}{2}}{2(x - 1)} = \lim_{x \to 1} \frac{1}{4} = -\frac{1}{4}. \] 
Final Answer: \( \boxed{-\frac{1}{4}} \)