limit $f(x)=((x-7)(x+4))^{1/2}$ I have to calculate the Asymptotes in the infinity(and minus infinity) for this function:
$f(x)=((x-7)(x+4))^{1/2}$
I know that
$\lim_{x \to\infty} f(x)/x= 1$
And I get into trouble with:
$\lim_{x \to\infty} f(x)-x$
which is
$\lim_{x \to\infty} ((x-7)(x+4))^{1/2}-x = \lim_{x \to\infty} \sqrt{x^2-3x-28}-x$
wolfram alpha says it is $-3/2$ but i don't get why.... please help me with that, thanks

$$
\sqrt{x^2-3x-28} - x = \frac{(\sqrt{x^2-3x-28} - x)(\sqrt{x^2-3x-28} + x)}{(\sqrt{x^2-3x-28} + x)}\\
= \frac{x^2-3x-28-x^2}{(\sqrt{x^2-3x-28} + x)}
= \frac{-3x-28}{x(\sqrt{1-3/x-28/x^2} + 1)} \to \frac {-3}{2}
$$
