Integral limit

For all $x \in \mathbb{R}$ we have that $\cos(x) = \displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(2n)!} = 1 - \dfrac{x^{2}}{2!} + \dfrac{x^{4}}{4!} - O(x^{6})$

and $e^{x} = \displaystyle\sum_{n=0}^{\infty} \dfrac{x^{n}}{n!} = 1 + x + \dfrac{x^{2}}{2!} + \dfrac{x^{3}}{3!} + O(x^{4})$ .

Thus $1 - \cos(x) = \dfrac{x^{2}}{2} - \dfrac{x^{4}}{24} + O(x^{6}) = \dfrac{x^{2}}{2}\left(1 - \dfrac{x^{2}}{12} + O(x^{4}) \right)$ and

$\cos(x) - e^{x} = \left(1 - \dfrac{x^{2}}{2!} + \dfrac{x^{4}}{4!} - O(x^{6}) \right) - \left(1 + x + \dfrac{x^{2}}{2!} + \dfrac{x^{3}}{3!} + \dfrac{x^{4}}{4!} + O(x^{5}) \right) =$

$- x - x^{2} - \dfrac{x^{3}}{6} - O(x^{5}) = -x(1 + x + O(x^{2})).$

Thus $\dfrac{(1 - \cos(x))(\cos(x) - e^{x})}{x^{n}} = \dfrac{-x^{3}(1 + x + O(x^{2}))}{2x^{n}} = -\dfrac{1}{2}x^{3 - n}(1 + x + O(x^{2})).$

For $n \lt 3$ the given limit will then be $0,$ and for $n \gt 3$ the limit will not exist. So only for $n = \boxed{3}$ will the limit be a finite non-zero number, namely $-\dfrac{1}{2}.$

1-cosx has "quadratic" power, and cosx-e^x has "linear" power, therefore, in order for the limit to exist, the denominator should have at least "cubic" power. What an awkward solution... but for this type of question where finding the minimum value of n is required, real calculation is not necessary, 1-cosx has "quadratic" power because it has non-zero limit iff n=2 because 1-cosx=2sin^2(x/2) and same logic is applied for the "linear" power.