why is $\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}}$=-3?

Question

Exercise taken from here: https://mooculus.osu.edu/textbook/mooculus.pdf (page 42, "Exercises for Section 2.2", exercise 4).

Why is $\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}}$=-3*? I always find 3 as the solution. I tried two approaches:

Approach 1: $$ \lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} = \lim_{x\to -\infty} \frac{3x+7}{x} = \lim_{x\to -\infty} \frac{3x}{x}+\lim_{x\to -\infty} \frac{7}{x}\\ = \lim_{x\to -\infty} 3+\lim_{x\to -\infty} \frac{7}{x} = 3 + 0 = 3 $$

Approach 2: $$ \lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} = \lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} \times \frac{\frac{1}{x}}{\frac{1}{x}} = \lim_{x\to -\infty} \frac{3+\frac{7}{x}}{\frac{\sqrt{x^2}}{x}}\\ = \lim_{x\to -\infty} \frac{3+\frac{7}{x}}{\frac{\sqrt{x^2}}{\sqrt{x^2}}} = \lim_{x\to -\infty} \frac{3+\frac{7}{x}}{\sqrt{\frac{x^2}{x^2}}}\\ = \frac{\lim_{x\to -\infty}3+\lim_{x\to -\infty}\frac{7}{x}}{\lim_{x\to -\infty}\sqrt{1}}\\ = \frac{3 + 0}{1} = 3 $$

* -3 is given as the answer by the textbook (cf. page 247) as well as wolfram|alpha

EDIT: I just re-read page 40 of the textbook and realized that I made a mistake in my approach 2. Instead of multiplying with $\frac{1}{x}$ I should have multiplied with $\frac{-1}{x}$ which is positive because as $x\to -\infty$, x is a negative number.

It thus reads:

$$ \lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} = \lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}} \times \frac{\frac{-1}{x}}{\frac{-1}{x}} = \lim_{x\to -\infty} \frac{-3+\frac{-7}{x}}{\frac{\sqrt{x^2}}{-x}}\\ = \lim_{x\to -\infty} \frac{-3+\frac{-7}{x}}{\frac{\sqrt{x^2}}{\sqrt{x^2}}} = \lim_{x\to -\infty} \frac{-3+\frac{-7}{x}}{\sqrt{\frac{x^2}{x^2}}}\\ = \frac{\lim_{x\to -\infty}-3+\lim_{x\to -\infty}\frac{-7}{x}}{\lim_{x\to -\infty}\sqrt{1}}\\ = \frac{-3 + 0}{1} = -3 $$

Answer 1

$\sqrt{x^2}=|x|$ for real $x$

For $x\le0, |x|=-x$

Answer 2

For $x \to -\infty$, we have that $\sqrt{x^2} = |x| = -x$.