Find $c$ and $n$ such that $\frac{x^3 \arctan x}{x^4 + \cos x +3} \sim cx^n$ as $x \to 0$

Question

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 $x^{n-4} = x^{-2}$ $\implies n=2$

$$-c \frac{1}{2} =1 \implies c=-2$$

Answer 1

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 $c=\frac14$.

Answer 2

It should be obvious that $c\neq 0$ otherwise the definition of $\sim$ is in trouble. Next we can note that the given condition implies $$\lim_{x\to 0}\frac{cx^n(3+\cos x+x^4)}{x^3\arctan x} =1$$ Using the limit $\lim_{x\to 0} (1/x)\arctan x=1$ we can see that the above condition is equivalent to $$4c\lim_{x\to 0}x^{n-4}=1$$ Now it should be obvious that $n=4$ as $n<4$ makes the limit infinite and $n>4$ makes the limit $0$ and then $c=1/4$.