Can You Limit It?

Relevant wiki: Maclaurin Series

$\begin{aligned} \mathfrak L & = \lim_{x \to 0} \frac{(\color{#3D99F6}{\cos x}-1)(\color{#3D99F6}{\cos x} - \color{#3D99F6}{e^x})}{x^n} \quad \quad \small \color{#3D99F6}{\text{Using Maclaurin series}} \\ & = \lim_{x \to 0} \frac{\left(\color{#3D99F6}{1-\frac{x^2}{2!} + \frac{x^4}{4!}-\frac{x^6}{6!}+\cdots}-1\right) \left( \color{#3D99F6}{1-\frac{x^2}{2!} + \frac{x^4}{4!}-\frac{x^6}{6!}+\cdots} - \color{#3D99F6}{1 - x - \frac{x^2}{2!}-\frac{x^3}{3!}+\cdots} \right)}{x^n} \\ & = \lim_{x \to 0} \frac{\left(-\frac{x^2}{2!} + \frac{x^4}{4!}-\frac{x^6}{6!}+\cdots \right) \left(-x-x^2 - \frac{x^3}{3!} - \frac{x^5}{5!}-\cdots \right)}{x^n} \\ & = \lim_{x \to 0} \frac{x^\color{#D61F06}{3} \left(-\frac{1}{2!} + \frac{x^2}{4!}-\frac{x^4}{6!}+\cdots \right) \left(-1-x - \frac{x^2}{3!} - \frac{x^4}{5!}-\cdots \right)}{x^\color{#D61F06}{n}} \quad \quad \small \color{#D61F06}{\text{Setting }n=\boxed{3}} \\ & = \lim_{x \to 0} \left(-\frac{1}{2!} + \frac{x^2}{4!}-\frac{x^4}{6!}+\cdots \right) \left(-1-x - \frac{x^2}{3!} - \frac{x^4}{5!}-\cdots \right) \\ & = \frac{1}{2} \ne 0 \end{aligned}$

The answer is $n = \boxed{3}$ .

Using trigonometric half angle formula

cosx-1 = -2sin^2(x/2)

Using popular limit x tends to zero sinx / x = 1

Ignoring that constant because we have to find n

cosx-e^x / x^(n-2)

Applying L Hospitals rule for 0/0 Form

-sinx-e^x / (n-2)x^(n-3)

Now this is not 0/0 form because numerator is -1 hence for limit to be finite and non zero n-3 = 0

Therefore n = 3

The value of n cant be less than 3 as we might end up getting 0 as limit