Answer
Not necessarily.
Example given below.
Work Step by Step
The numerator, $f(x)$ may have a factor that cancels the term in $g(x)$ that yields 0 in the denominator.
In this case, a finite limit exists and there is no vertical asymptote.
Example: $F(x)=\displaystyle \frac{(x-1)(x+2)}{(x-1)(x^{2}+1)}$ is such a function $(g(1)=0)$.
Cancelling the common factor, we have
$\displaystyle \lim_{x\rightarrow 1}F(x)=\lim_{x\rightarrow 1}[\frac{(x+2)}{x^{2}+1}]=\frac{3}{2}$
(None of the one sided limits is is $\pm\infty$,
so there is no vertical asymptote at $x=1)$
