There's no limit to it

We can use the following Maclaurin expansions:

$\cos(x)=1-\frac12x^2+\frac1{24}x^4+\mathcal{O}(x^6)$

$e^x=1+x+\frac12x^2+\frac16x^3+\frac1{24}x^4+\mathcal{O}(x^5)$

where $\mathcal{O}$ indicates the big O-notation . Plugging these into the limit formula and simplifying gives us:

$\cos^2(x)-\cos(x)-e^x\cos(x)+e^x-\frac{x^3}2 = \frac12x^4+\mathcal{O}(x^5)$ .

Hence for $n<4$ the limit would be equal to $0$ , whereas for $n=4$ the limit is equal to $\frac12$ . Therefore the answer is $n=\boxed{4}$ .