<s>[INST] When computing $\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}$, I don't get -2. This limit:

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

is supposed to be equal to -2. My textbook and Wolfram Alpha both state that. However, I can't seem to get same exact result.
Here's what I tried to do:

$$\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}} = \lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}} \cdot \frac{1 \over x}{1 \over x} = \lim_{x\to -\infty}\frac{\frac{x}{x}\cdot(2+\frac7x)}{\sqrt{\frac{x^2}{x^2}\cdot(1+\frac{2}{x}-\frac{1}{x^2})}} = \frac{1\cdot(2+0)}{\sqrt{1\cdot(1+0-0)}} =\frac{2}{\sqrt{1}}=2$$

Where did I make a mistake?
 [/INST] We have that : $\frac{2x+7}{\sqrt{x^2+2x-1}}=\frac{x(2+\frac{7}{x})}{\sqrt{x^2(1+\frac{2}{x}-\frac{1}{x^2})}}=\frac{x(2+\frac{7}{x})}{(\left | x \right |)\sqrt{1+\frac{2}{x}-\frac{1}{x^2}}}$. Now use the fact that $\left | x \right |=-x$ for negative $x$ and youre done.
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