Question:medium
\(\lim_{x \to -2} \frac{\sin^{-1}(x + 2)}{x^2 + 2x}\) is equal to
\(\lim_{x \to -2} \frac{\sin^{-1}(x + 2)}{x^2 + 2x}\) is equal to
Show Hint
\(\lim_{u \to 0} \frac{\sin^{-1}u}{u} = 1\) and \(\lim_{u \to 0} \frac{\tan^{-1}u}{u} = 1\).
Updated On: Apr 7, 2026
- 0
- \(\infty\)
- \(-1/2\)
- None of these
Show Solution
The Correct Option is C
Solution and Explanation
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