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

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