Asymptotes in $\frac{3x+\sin x}{2x-\cos x}$ I am asked to determine whether the following function:
 $$\frac{3x+\sin x}{2x-\cos x}$$
has a horizontal, vertical, both horizontal and vertical, or oblique asymptote(s).
I can easily find the horizontal one as the limit of this function as $x$ approaches $+\infty$ is a finite number, $\frac{3}{2}$. 
However, in order to determine the existence of a vertical asymptote, I should evaluate the limit of a problematic area in the function, which would be the value of $x$ such that the function is undefined. That value is surely $x :2x -\cos x = 0$ but I don't know how to find that value. 
Hence, I cannot determine the existence of a vertical asymptote because I cannot evalue the corresponding limit.
Any help?
 A: You are correct that you need to investigate $2x-\cos x=0$, or $\cos x=2x$.  But they don't ask you to identify the vertical asymptote, just identify that one exists.  If that equation has a solution (and it does have exactly one solution), and that solution isn't also a zero of the numerator (and it couldn't possibly be), then the graph has a vertical asymptote even if you don't know the precise value.
A: Hint: You want to find all $x$ such that $2x = \cos(x)$.
For $x > \frac{1}{2}$ we have $2x > 1$ and since $\cos(x) \in [-1,1]$ for all $x \in \mathbb{R}$ we can rule this out.
By symmetry we obtain that the relevant points have to be contained in $\left[-\frac{1}{2}, \frac{1}{2}\right]$.
Furthermore, $\cos(x) > 0 > 2x$ for all $- \frac{1}{2} < x < 0$. So we only have to check $\left[0, \frac{1}{2}\right]$.
A: You also need that the horizontal asymptote as $x\to -\infty$ is $\dfrac{3}{2}.$ Vertical asymptotes are solutions to the equation $2x-\cos x=0\Rightarrow\cos x=2x.$ Let $f(x)=2x-\cos x.$ Since $f(x)$ is continuous (this can be shown) and $f(-1) <0$ and $f(1)>0,$ by the IVT, $\exists x_0\in\mathbb{R}$ such that $f(x_0)=0.$ We have that $f'(x) = 2+\sin x.$ So $f'(x)>0\;\forall x\in\mathbb{R}.$ Now, if it had more than one zero, then by Rolle's theorem, there would exist an $x_1$ such that $f'(x_1)=0,$ which is a contradiction. So it only has one zero. And this zero is approximately at $x=0.45018.$
