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.

$$\sqrt{x^2+3x}-\sqrt{x^2+1}=\frac{3x-1}{\sqrt{x^2+3x}+\sqrt{x^2+1}}\cdot\frac{\frac1{-x}}{\frac1{-x}}=$$
$$=\frac{-3+\frac1x}{\sqrt{1+\frac3x}+\sqrt{1+\frac1{x^2}}}\xrightarrow[x\to-\infty]{}-\frac32$$
You need the $\;-x\;$ in order to make it positive and then you can take it into the square root.
