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.

$$\lim_{x\to-\infty}\sqrt{x^2+3x}-\sqrt{x^2+1}=\lim_{x\to-\infty}\left(\sqrt{x^2+3x}-\sqrt{x^2+1}\right)\cdot 1$$
$$=\lim_{x\to-\infty}\sqrt{x^2+3x}-\sqrt{x^2+1}\cdot\frac{\sqrt{x^2+3x}+\sqrt{x^2+1}}{\sqrt{x^2+3x}+\sqrt{x^2+1}}$$
$$=\lim_{x\to-\infty}\frac{\sqrt{x^2+3x}^2-\sqrt{x^2+1}^2}{\sqrt{x^2+3x}+\sqrt{x^2+1}}=\lim_{x\to-\infty}\frac{x^2+3x-(x^2+1)}{\sqrt{x^2\left(1+\frac{3}{x}\right)}+\sqrt{x^2\left(1+\frac{1}{x}\right)}}$$
$$=\lim_{x\to-\infty}\frac{3x-1}{2x}=-\frac{3}{2}$$
