Problem Statement
Find the first three non-zero terms in the Maclaurin series expansion of $f(x) = e^x \cos x$. Hence, find $\lim_{x \to 0} \frac{e^x \cos x - 1 - x}{x^3}$.
Verified Solution & Marking Scheme
Write Known Maclaurin Series
$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots$
$\cos x = 1 - \frac{x^2}{2} + \dots$
Multiply Series up to x³
$e^x \cos x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots\right) \left(1 - \frac{x^2}{2} + \dots\right)$
Term $x^0$: $1 \times 1 = 1$.
Term $x^1$: $x \times 1 = x$.
Term $x^2$: $1\left(-\frac{x^2}{2}\right) + \frac{x^2}{2}(1) = 0$.
Term $x^3$: $x\left(-\frac{x^2}{2}\right) + \frac{x^3}{6}(1) = -\frac{x^3}{2} + \frac{x^3}{6} = -\frac{2x^3}{6} = -\frac{x^3}{3}$.
Thus:
$e^x \cos x = 1 + x - \frac{x^3}{3} + \dots$
Evaluate Limit
$\lim_{x \to 0} \frac{\left(1 + x - \frac{x^3}{3} + O(x^4)\right) - 1 - x}{x^3} = \lim_{x \to 0} \frac{-\frac{x^3}{3} + O(x^4)}{x^3} = -\frac{1}{3}$