Question

Evaluate the following limit: $ \lim_{x \to 0} \frac{1 + \cos(4x)}{\tan(x)} $

• A. 2
• B. 1
• C. 0
• D. 4

Hint:

For small angle approximations, remember that \( \cos(x) \approx 1 - \frac{x^2}{2} \) and \( \tan(x) \approx x \). These approximations are very useful for solving limits involving trigonometric functions as \( x \to 0 \).

Answer 1

The Correct Option is B

We need to evaluate the limit as \( x \to 0 \) of the given expression: \[ \lim_{x \to 0} \frac{1 + \cos(4x)}{\tan(x)} \] First, recall the small angle approximations: - \( \cos(4x) \approx 1 - \frac{(4x)^2}{2} \) as \( x \to 0 \), - \( \tan(x) \approx x \) as \( x \to 0 \).
Thus, we can rewrite the expression: \[ \frac{1 + \cos(4x)}{\tan(x)} \approx \frac{1 + \left( 1 - \frac{16x^2}{2} \right)}{x} = \frac{2 - 8x^2}{x} \] As \( x \to 0 \), the term \( -8x^2 \) vanishes, and we are left with: \[ \frac{2}{x} \]
Thus, the limit evaluates to: \[ 1 \]