The value of $\lim_{x \to 0} \frac{(1 + \sin 2x)^{1/\cos x}}{x \sin 8x}$ is equal to
Step-by-Step Solution
Key Concept: Use trigonometric identities to simplify the numerator and apply standard limits $\lim_{x \to 0} \frac{\sin x}{x} = 1$
We have $\lim_{x \to 0} \frac{(1-\cos 2x)(1+\sin x)}{x\sin 4x}$. Using the identity $(1-\cos 2x) = 2\sin^2 x$ and standard limits, we rewrite as $\lim_{x \to 0} \frac{2\sin^2 x(1+\sin x)}{x \cdot 4\sin x \cos x} = \lim_{x \to 0} \frac{2\sin x(1+\sin x)}{4x\cos x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{2(1+\sin x)}{4\cos x} = 1 \cdot \frac{2}{4} = \frac{1}{2}$.
Correct Answer: 2