Question:
$\lim_{x\rightarrow 0}\frac{\sin x}{2\sqrt{2}\sin\frac{x}{\sqrt{2}}} =$
$\lim_{x\rightarrow 0}\frac{\sin x}{2\sqrt{2}\sin\frac{x}{\sqrt{2}}} =$
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Small angle approximation $\sin \theta \approx \theta$ makes these limit problems solvable at a glance.
Updated On: Apr 28, 2026
- $\sqrt{2}$
- $2\sqrt{2}$
- $\frac{1}{\sqrt{2}}$
- $\frac{1}{2\sqrt{2}}$
- $\frac{1}{2}$
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The Correct Option is
Solution and Explanation
Step 1: Concept
Use the standard limit $\lim_{\theta\rightarrow0} \frac{\sin \theta}{\theta} = 1$.
Step 2: Analysis
Multiply and divide by variables to match the limits: $\lim_{x\rightarrow0} \frac{(\frac{\sin x}{x}) \cdot x}{2\sqrt{2} (\frac{\sin(x/\sqrt{2})}{x/\sqrt{2}}) \cdot (x/\sqrt{2})}$
Step 3: Calculation
Value $= \frac{1 \cdot x}{2\sqrt{2} \cdot 1 \cdot (x/\sqrt{2})} = \frac{x}{2x} = \frac{1}{2}$. Final Answer: (E)
Use the standard limit $\lim_{\theta\rightarrow0} \frac{\sin \theta}{\theta} = 1$.
Step 2: Analysis
Multiply and divide by variables to match the limits: $\lim_{x\rightarrow0} \frac{(\frac{\sin x}{x}) \cdot x}{2\sqrt{2} (\frac{\sin(x/\sqrt{2})}{x/\sqrt{2}}) \cdot (x/\sqrt{2})}$
Step 3: Calculation
Value $= \frac{1 \cdot x}{2\sqrt{2} \cdot 1 \cdot (x/\sqrt{2})} = \frac{x}{2x} = \frac{1}{2}$. Final Answer: (E)
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