Evaluate the limit: $$\lim_{x \to 0} \frac{1 - \cos(4x)}{x^2}$$

(A) 8

(B) 4

(C) 16

(D) 2

Answer Key:

A

### Solution: Using Taylor series: $\cos(4x) \approx 1 - \frac{(4x)^2}{2} = 1 - 8x^2$. Thus $\frac{1 - (1 - 8x^2)}{x^2} = 8$. Correct: A.

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