Question

$\lim_{x\rightarrow 0}\frac{\sin x}{2\sqrt{2}\sin\frac{x}{\sqrt{2}}} =$

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2\sqrt{2}}$
• E. $\frac{1}{2}$

Hint:

Small angle approximation $\sin \theta \approx \theta$ makes these limit problems solvable at a glance.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We need to evaluate a limit of a trigonometric function as x approaches 0. This limit is in the indeterminate form \( \frac{0}{0} \), so we can use the standard trigonometric limit \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \).
Step 2: Key Formula or Approach:
We will rearrange the expression to make use of the identity \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \). This involves multiplying and dividing by appropriate terms.
Step 3: Detailed Explanation based on OCR interpretation:
The given limit is: \[ L = \lim_{x \to 0} \frac{\sin x}{2\sqrt{2}\sin\left(\frac{x}{\sqrt{2}}\right)} \] Let's manipulate the numerator and denominator to create the \( \frac{\sin \theta}{\theta} \) form. \[ L = \frac{1}{2\sqrt{2}} \lim_{x \to 0} \frac{\sin x}{\sin\left(\frac{x}{\sqrt{2}}\right)} \] \[ L = \frac{1}{2\sqrt{2}} \lim_{x \to 0} \left( \frac{\sin x}{x} \cdot \frac{\frac{x}{\sqrt{2}}}{\sin\left(\frac{x}{\sqrt{2}}\right)} \cdot \frac{x}{x/\sqrt{2}} \right) \] As \( x \to 0 \), we have \( \frac{x}{\sqrt{2}} \to 0 \). We can separate the limits: \[ L = \frac{1}{2\sqrt{2}} \left( \lim_{x \to 0} \frac{\sin x}{x} \right) \cdot \left( \lim_{x \to 0} \frac{\frac{x}{\sqrt{2}}}{\sin\left(\frac{x}{\sqrt{2}}\right)} \right) \cdot \left( \lim_{x \to 0} \frac{x}{x/\sqrt{2}} \right) \] Using the standard limit, the first two parts are equal to 1. \[ L = \frac{1}{2\sqrt{2}} \cdot (1) \cdot (1) \cdot \left( \lim_{x \to 0} \sqrt{2} \right) \] \[ L = \frac{1}{2\sqrt{2}} \cdot \sqrt{2} = \frac{1}{2} \] This calculation matches the correct answer. The OCR appears to be correct after all. Step 4: Final Answer:
The value of the limit is \( \frac{1}{2} \).