Question:
Evaluate
\[
\lim_{x \to 0} \sqrt{\frac{x - \sin x}{x + \sin^2 x}}
\]
Evaluate
\[ \lim_{x \to 0} \sqrt{\frac{x - \sin x}{x + \sin^2 x}} \]
\[ \lim_{x \to 0} \sqrt{\frac{x - \sin x}{x + \sin^2 x}} \]
Show Hint
Use series expansions for limits involving trig functions.
Updated On: Mar 23, 2026
- \(1\)
- \(0\)
- \(\infty\)
- None of these
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Use expansions:
\[ \sin x = x - \frac{x^3}{6} + \cdots \]
Step 2:
\[ x - \sin x \sim \frac{x^3}{6}, \quad x + \sin^2 x \sim x + x^2 \]
Step 3:
\[ \frac{x - \sin x}{x + \sin^2 x} \sim \frac{x^3/6}{x} = \frac{x^2}{6} \]
Step 4:
\[ \sqrt{\frac{x^2}{6}} \to 0 \quad \text{as } x \to 0 \]
\[ \sin x = x - \frac{x^3}{6} + \cdots \]
Step 2:
\[ x - \sin x \sim \frac{x^3}{6}, \quad x + \sin^2 x \sim x + x^2 \]
Step 3:
\[ \frac{x - \sin x}{x + \sin^2 x} \sim \frac{x^3/6}{x} = \frac{x^2}{6} \]
Step 4:
\[ \sqrt{\frac{x^2}{6}} \to 0 \quad \text{as } x \to 0 \]
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