Question

Evaluate the limit: \[ \lim_{x \to 0} \frac{1 - \cos 4x}{2 \sin^2 x + x \tan 7x} \]

• A. \( \frac{8}{9} \)
• B. 0
• C. \( \frac{9}{8} \)
• D. \( \infty \)

Hint:

When dealing with small angle approximations, expand trigonometric functions using their series expansions and simplify the resulting expressions.

Answer 1

The Correct Option is A

We use series expansions of trigonometric functions around 0: \[ \cos 4x \approx 1 - 8x^2 \quad {and} \quad \sin x \approx x \quad {and} \quad \tan 7x \approx 7x \] Substituting these approximations into the expression: \[ \lim_{x \to 0} \frac{1 - (1 - 8x^2)}{2x^2 + x(7x)} = \frac{8x^2}{2x^2 + 7x^2} = \frac{8}{9} \]