Is $x=-1$ a Vertical tangent or Asymptote to the curve $f(x)=\frac{(x-1)^2}{x+1}$
I have a little confusion about the difference between VT and Asymptote
we have
$$f'(x)=1-\frac{4}{(x+1)^2}$$ so
$$ \lim _{x \to -1^+} f'(x) \to - \infty$$ and
$$ \lim _{x \to -1^-} f'(x) \to - \infty$$ and
So at $x=-1$ $f(x)$ has VT.
Also by definition of asymptote we have
$$ \lim_{x \to -1^+} f(x) \to \infty$$ and
$$ \lim_{x \to -1^-} f(x) \to -\infty$$
hence $x=-1$ is Vertical Asymptote.
so is this correct?