<s>[INST] Is this limit at infinity negative or positive 1 
$\lim_{x \to -∞} \frac{x}{\sqrt{x^2+4}}$

I solved it:
$\lim_{x \to -∞} \frac{x}{\sqrt{x^2+4}} = \frac{x}{\sqrt{x^2}} = \frac{x}{x} = 1 , -1$ ?
Since its $x/x$, a negative infinity number divided by another negative infinity number is a positive infinity. But I am told that this limit is $-1$
 [/INST] Note that
$$
\frac{x}{\sqrt{x^2+4}} = \frac{x}{\sqrt{x^2(1+4x^{-2}})} = \frac{x}{|x|}\frac{1}{\sqrt{1+4x^{-2}}}
$$
so
$$
\lim_{x\to -\infty}\frac{x}{\sqrt{x^2+4}} = \lim_{x\to -\infty}\frac{x}{|x|} \times\lim_{x\to -\infty}\frac{1}{\sqrt{1+4x^{-2}}} = -1\times 1 = -1
$$
since
$$
\frac{x}{|x|} = -1
$$
for $x<0$ and $(1+4x^{-2})^{-1/2}\to 1$ as $x\to -\infty$.
</s>