What is the mistake in my reasoning in calculating limit where $x \to -\infty$ I had this question in an exam today:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x}
$$
My question is: I have tried to solve it and I also found two other methods to solve it, one in my textbook and one from somebody else. All three methods give different answers, and I can't see the issue with either of the methods which are not mine. I need help figuring out what the invalid operation(s) in each of the wrong methods are, and which 2 (or 3) methods are wrong.
I didn't solve it in the exam, but afterwards tried this:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \Bigg(\frac{\sqrt{x^2+4x}}{2x}\Bigg)  - 1 \\
= \; \frac{1}{2}\lim \limits_{x \, \to \,-\infty} \; \Bigg(\frac{\sqrt{x^2+4x}}{x}\Bigg)  - 1 \\
= \; \frac{1}{2}\lim \limits_{x \, \to \,-\infty} \; \Big(\sqrt{x^2+4x}\cdot \frac{1}{x}\Big)  - 1 \\
= \; \frac{1}{2} \; \Big(\lim \limits_{x \, \to \,-\infty}\sqrt{x^2+4x} \, \cdot \, \lim \limits_{x \, \to \,-\infty}\frac{1}{x}\Big)  - 1 \\
= \; \frac{1}{2} \; \Bigg(\sqrt{\lim \limits_{x \, \to \,-\infty}x^2+4x}\,\cdot \, \lim \limits_{x \, \to \,-\infty}\frac{1}{x}\Bigg)  - 1 \\
= \; \frac{1}{2} \; \Bigg(\sqrt{\lim \limits_{x \, \to \,-\infty}x^2+4x}\,\cdot \, 0\Bigg)  - 1 \\
= \; -1 \\
$$
I understand my mistake: I multiply an infinite amount with 0 and say it is equal to 0 in my last step.
However, our study guide has that exact example with the following solution:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2(1+\frac{4}{x})}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{|x|\sqrt{(1+\frac{4}{x})}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{-x\sqrt{(1+\frac{4}{x})}-2x}{2x} \; ; \; (x < 0) \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{-x(\sqrt{(1+\frac{4}{x})}+2)}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{-(\sqrt{(1+\frac{4}{x})}+2)}{2} \\
= \; \frac{-\Bigg(\sqrt{\lim \limits_{x \, \to \,-\infty} \; (1+\frac{4}{x})}+2\Bigg)}{2} \\
= \; \frac{-(\sqrt{1+0}+2)}{2} \\
= \; \frac{-(\sqrt{1}+2)}{2} \\
= \; \frac{-(1+2)}{2} \\
= \; -\frac{3}{2}
$$
To complicate matters even more: someone else solved the problem like this and got a third answer:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}}{2x} - 1 \\
= \; \lim \limits_{x \, \to \,-\infty} \; \sqrt{\frac{x^2+4x}{4x^2}} - 1 \\
= \; \lim \limits_{x \, \to \,-\infty} \; \sqrt{\frac{1}{4} + \frac{1}{x}} - 1 \\
= \; \; \sqrt{\lim \limits_{x \, \to \,-\infty} \; \Big(\frac{1}{4} + \frac{1}{x}\Big)} - 1 \\
= \; \sqrt{\lim \limits_{x \, \to \,-\infty} \; \frac{1}{4} + \lim \limits_{x \, \to \,-\infty} \; \frac{1}{x}} - 1 \\
= \; \sqrt{\frac{1}{4} + 0} - 1 \\
= \; \sqrt{\frac{1}{4}} - 1 \\
= \; \frac{1}{2} - 1 \\
= \; -\frac{1}{2}
$$
 A: Exact explanations were made about the mistake you made in the comments. You missed that $x$ is negative.
You can observe this simple fact:
$$\begin{align}x=\begin{cases}-\sqrt{x^2},~\text{if}~ x<0\\\sqrt{x^2}, ~\text{if}~x>0\end{cases}\end{align}$$
This immediately reveals your mistake.
I would calculate this limit as follows:
$$\begin{align}\lim \limits_{x\to -\infty} \frac{\sqrt{x^2+4x}-2x}{2x}&=-\lim_{x \to\infty} \frac{\sqrt{x^2-4x}+2x}{2x}\\
&=-\lim_{x \to\infty} \left(\sqrt{\frac{x^2}{4x^2}-\frac{4x}{4x^2}}+1\right)\\
&=-\lim_{x \to\infty} \left(\sqrt{\frac 14-\frac{1}{x}}+1\right)\\
&=-\left(\frac 12+1\right)\\
&=-\frac 32.\end{align}$$
A: As noticed the second method is the correct one, to check the result we can make the limit simpler at positive $\infty$ by $u=-x\to \infty$, then
$$\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x} =\lim \limits_{u \, \to \,\infty} \; \frac{\sqrt{u^2-4u}+2u}{-2u}=\lim \limits_{u \, \to \,\infty} \; \frac{\sqrt{1-\frac4 u}+2}{-2}=-\frac 3 2$$
