Problem Statement
Consider the standard Maclaurin series $\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \dots$
**(a)** [3 marks] Find the Maclaurin series for $f(x) = \cos(x^2)$ up to and including the term in $x^8$.
**(b)** [4 marks] Hence, determine the exact value of $\lim_{x \to 0} \frac{1 - \cos(x^2)}{x^4}$.
Verified Solution & Marking Scheme
Part (a) [M1]
Substitute $u = x^2$ into the series: $\cos(x^2) = 1 - \frac{(x^2)^2}{2!} + \frac{(x^2)^4}{4!} - \dots$
Part (a) [A1]
Simplify exponents: $\cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \dots$
Part (a) [A1]
Correct non-zero terms up to $x^8$ given explicitly
Part (b) [M1]
Substitute series into numerator: $1 - \cos(x^2) = 1 - \left(1 - \frac{x^4}{2} + \frac{x^8}{24} - \dots\right) = \frac{x^4}{2} - \frac{x^8}{24} + \dots$
Part (b) [M1]
Divide through by $x^4$: $\frac{1 - \cos(x^2)}{x^4} = \frac{1}{2} - \frac{x^4}{24} + O(x^8)$
Part (b) [R1]
Evaluate limit as $x \to 0$: all terms with powers of $x$ vanish
Part (b) [A1]
Limit is equal to $\frac{1}{2}$