How to think about negative infinity in this limit $\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$ 
Question:
calculate:
$$\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$$

Attempt at a solution:
This can be written as:
$$\lim_{x \to -\infty} \frac{3 + \frac{1}{x}}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}}}$$
Here we can clearly see that if x would go to $+\infty$ the limit would converge towards $\frac{3}{2}$. But what happens when x goes to $-\infty$.
From the expression above it would seem that the answer would still be $\frac{3}{2}$. My textbook says it would be $- \frac{3}{2}$ and I can't understand why.
I am not supposed to use l'Hospital's rule for this exercise.
 A: In fact:
$\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$
$ = \lim_{x \to -\infty} \frac{x*(3 + \frac{1}{x})}{(-x)(\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}})}$
$ = \lim_{x \to -\infty} \frac{-3 - \frac{1}{x}}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}}} = -\frac{3}{2}$
A: The reason your sign has changed from what it should be, is you illegally pulled something out of the square roots on the denominator.
$$\sqrt{a^2b}=|a|\sqrt{b}$$
the absolute value sign being essential.
A: $$\sqrt{x^2+3x}-\sqrt{x^2+1}=\frac{3x-1}{\sqrt{x^2+3x}+\sqrt{x^2+1}}\cdot\frac{\frac1{-x}}{\frac1{-x}}=$$
$$=\frac{-3+\frac1x}{\sqrt{1+\frac3x}+\sqrt{1+\frac1{x^2}}}\xrightarrow[x\to-\infty]{}-\frac32$$
You need the $\;-x\;$ in order to make it positive and then you can take it into the square root.
A: $$\lim_{x\to-\infty}\sqrt{x^2+3x}-\sqrt{x^2+1}=\lim_{x\to-\infty}\left(\sqrt{x^2+3x}-\sqrt{x^2+1}\right)\cdot 1$$
$$=\lim_{x\to-\infty}\sqrt{x^2+3x}-\sqrt{x^2+1}\cdot\frac{\sqrt{x^2+3x}+\sqrt{x^2+1}}{\sqrt{x^2+3x}+\sqrt{x^2+1}}$$
$$=\lim_{x\to-\infty}\frac{\sqrt{x^2+3x}^2-\sqrt{x^2+1}^2}{\sqrt{x^2+3x}+\sqrt{x^2+1}}=\lim_{x\to-\infty}\frac{x^2+3x-(x^2+1)}{\sqrt{x^2\left(1+\frac{3}{x}\right)}+\sqrt{x^2\left(1+\frac{1}{x}\right)}}$$
$$=\lim_{x\to-\infty}\frac{3x-1}{2x}=-\frac{3}{2}$$
