Discussion
\(\lim_{x \to 0} \frac{\sin x}{2\sqrt{2}\sin\frac{x}{\sqrt{2}}} =\)
Answer: E
The limit is \(\frac{1}{2}\).
Use \(\lim_{t \to 0}\frac{\sin t}{t} = 1\) for both sines, writing each over its own angle.
\[ \begin{aligned} \frac{\sin x}{2\sqrt{2}\sin\frac{x}{\sqrt{2}}} &= \frac{\frac{\sin x}{x} \cdot x}{2\sqrt{2}\cdot\frac{\sin(x/\sqrt{2})}{x/\sqrt{2}}\cdot\frac{x}{\sqrt{2}}} \\ &\to \frac{x}{2x} = \frac{1}{2} \end{aligned} \]
The \(\sqrt{2}\) in the angle cancels the \(\sqrt{2}\) in the coefficient, leaving \(\frac{1}{2}\). \(\frac{1}{2\sqrt{2}}\) ignores the factor \(\frac{1}{\sqrt{2}}\) in the angle, and \(\frac{1}{\sqrt{2}}\) uses it twice.
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