$\lim_{x\to \infty}\frac{x+\cos x}{x - \cos x}$ Here $f(x) = x + \cos x$ and $g(x)= x- \cos x$
We can't apply L'Hopital's Rule since $\forall c > 0\quad \exists x_c \in (c, + \infty): g'(x) = 1 + \sin x = 0$.
But I don't know how to proceed. 
My attempt:
Because $\cos x$ is bounded I was taking an arbitrary number $A >0$ that bounds $\cos x$ from above.
From there I have
$$\lim_{x\to \infty}\frac{x+\cos x}{x - \cos x} = \lim_{x\to \infty}\frac{x+A}{x - A} =1$$
But the answer is $0$. What am I doing wrong?

For $x \gt 1$, you have $\dfrac{x+\cos x}{x - \cos x}=1 + \dfrac{2\cos x}{x - \cos x}$ 
Since $0 \le | \cos x | \le 1$ you have $-2 \le 2 \cos x  \le 2$ and $x-\cos x \ge x-1$ 
meaning $1 - \dfrac{2}{x - 1} \le \dfrac{x+\cos x}{x - \cos x} \le 1 + \dfrac{2}{x - 1}$ and thus a limit of $1$ as $x$ increases without limit
