Why isn't there a vertical asymptote at $x = -1$ for $f(x) = \frac{(x^3-1)(\ln(x+2))}{2x^3+3x^2-2x-3}$ Why isn't there a vertical asymptote at $x = -1$ for $f(x) = \frac{(x^3-1)(\ln(x+2))}{2x^3+3x^2-2x-3}$
I understand that there is a removable discontinuity at x = 1 since $x^3 - 1 = (x-1)(x^2+x+1)$ but I don't know how x = -1 is not a vertical asymptote.

hint
The denominator is
$$(2x+3)(x-1)(x+1) $$
and we use
$$\lim_{x\to-1}\frac {\ln (x+2)}{x+1}=$$
$$\lim_{X\to 0}\frac {\ln (X+1)}{X}=1$$
we find that
$$\lim_{x\to-1}f (x)=\frac {-2}{-2}=1$$
Thus, there is No vertical asymptote at $x=-1$ since the limit is finite.
