Find vertical asymptotes of function I have problem solve this task.
Find vertical asymptotes of this function:
$$f(x)=\frac{\sqrt{x^2-4}}{2x+1}$$
I found the domain:$$D\in(-\infty,2]\cup[2,+\infty)$$
I solved limes:
$$\lim\limits_{x\to-2^{-}}\frac{\sqrt{x^2-4}}{2x+1}=0$$
and
$$\lim\limits_{x\to2^{+}}\frac{\sqrt{x^2-4}}{2x+1}=0$$
And I concluded that there are no vertical asymptotes?
Is that right?

To find the largest possible domain, we want $x^2-4 \ge 0$ and $2x+1 \ne 0$, which corresponds to $$((-\infty, -2] \cup [2, \infty))-\{-\frac12\}=(-\infty, -2] \cup [2, \infty).$$
The function does not have a denominator that can take value $0$, there is no vertical asymptote. I have attached a plot of the function.

