Question

\(\lim_{x \to -2} \frac{\sin^{-1}(x + 2)}{x^2 + 2x}\) is equal to

• A. 0
• B. \(\infty\)
• C. \(-1/2\)
• D. None of these

Hint:

\(\lim_{u \to 0} \frac{\sin^{-1}u}{u} = 1\) and \(\lim_{u \to 0} \frac{\tan^{-1}u}{u} = 1\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Use substitution and limit formula \(\lim_{u \to 0} \frac{\sin^{-1}u}{u} = 1\).
Step 2: Detailed Explanation:
Let \(u = x + 2 \rightarrow\) as \(x \to -2\), \(u \to 0\)
\(x^2 + 2x = (u - 2)^2 + 2(u - 2) = u^2 - 4u + 4 + 2u - 4 = u^2 - 2u = u(u - 2)\)
Limit = \(\lim_{u \to 0} \frac{\sin^{-1}u}{u(u - 2)} = \lim_{u \to 0} \frac{\sin^{-1}u}{u} \cdot \frac{1}{u - 2} = 1 \cdot \frac{1}{-2} = -\frac{1}{2}\)
Step 3: Final Answer:
Limit = \(-1/2\).