Find $c$ and $n$ such that $\frac{x^3 \arctan x}{x^4 + \cos x +3} \sim cx^n$ as $x \to 0$ Where is my mistake in the below:
$$\frac{1}{c} \lim_{x \to 0} \frac{x^3 \arctan x}{x^{4+n} + x^n \cos x + 3x^n} = \frac{1}{c} \lim_{x \to 0} \frac{x^4}{x^{4+n}+x^n \cos x + 3x^n} \\ =\frac{1}{c} \lim_{x \to 0} \frac{1}{x^n + x^{n-4} \cos x + 3x^{n-4}}$$
Looking at it now it feels like it is the step that I am about to do, because it appears as though the denominator becomes smaller and smaller going to infinity. However, my reasoning was that since it is given that there can be an equivalence between $cx^n$ and our function, i.e. the limit is finite and goes to $1$ as $x \to 0$, then it also must be true that if we invert the fraction, then the limit is still finite and is $1/1=1$ as $x \to 0$. So, next step:
$$c \lim_{x \to 0} \left(x^n + x^{n-4}\cos x + 3x^{n-4} \right)$$
$$c \lim_{x \to 0} (x^{n-4}(\cos x -1))=1$$
$$-c \lim_{x \to 0} \left(x^{n-4} \frac{x^2}{2}\right) = 1$$
We need $x^{n-4} = x^{-2}$ $\implies n=2$
$$-c \frac{1}{2} =1 \implies c=-2$$

We have that
$$\frac{x^3 \arctan x}{x^4 + \cos x +3}\sim\frac{x^4}{x^4+4}\sim \frac14 x^4$$
and the guess is correct indeed
$$\frac{\frac{x^3 \arctan x}{x^4 + \cos x +3}}{\frac14 x^4}\to 1$$
As an alternative
$$\frac{\frac{x^3 \arctan x}{x^4 + \cos x +3}}{c x^n}=\frac1c\frac{x^3 \arctan x}{x^{4+n} + x^n\cos x +3x^n}=\frac1c\frac{x^4 +o(x^4)}{x^{4+n} + x^n+o(x^{n}) +3x^n}=$$$$=\frac1c\frac{1 +o(1)}{4x^{n-4}+o(x^{n-4})} \to 1$$
from which we obtain $n=4$ and $c=\frac14$.
