Problem Statement
Use Maclaurin series expansions to evaluate: $\lim_{x \to 0} \frac{\cos(2x) - 1 + 2x^2}{x^4}$
Verified Solution & Marking Scheme
State standard Maclaurin series for cos(u)
$\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \frac{u^6}{6!} + \dots$
Substitute $u = 2x$:
$\cos(2x) = 1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24} + O(x^6) = 1 - 2x^2 + \frac{16x^4}{24} + O(x^6) = 1 - 2x^2 + \frac{2}{3}x^4 + O(x^6)$
Substitute expansion into numerator
$\cos(2x) - 1 + 2x^2 = \left(1 - 2x^2 + \frac{2}{3}x^4 + O(x^6)\right) - 1 + 2x^2 = \frac{2}{3}x^4 + O(x^6)$
Divide by x⁴ and take limit as x → 0
$\lim_{x \to 0} \frac{\frac{2}{3}x^4 + O(x^6)}{x^4} = \lim_{x \to 0} \left( \frac{2}{3} + O(x^2) \right) = \frac{2}{3}$