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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?

Saraph
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3 Answers3

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The mistake at the second $=$ where you use $\frac{1}{x}=\sqrt{\frac{1}{x^2}}$ which is false when $x$ is negative. The correct formula is $\sqrt{x^2}=|x|$ which gives $\left|\frac{1}{x}\right|=\sqrt{\frac{1}{x^2}}$ in your case.

xavierm02
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    Oh, I think I know where problem is now. So if $x\to - \infty$, we can say that $-\frac{1}{x} = \sqrt{\frac{1}{x^2}}$, right? – Saraph Sep 25 '13 at 19:03
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    @Saraph : Yes :) – xavierm02 Sep 25 '13 at 19:08
  • Alright, thank you. Your answer made me realize where my brain fails to operate correctly. I'm a bit smarter now. ;) – Saraph Sep 25 '13 at 19:14
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    @Saraph : It's not your brain. It is a very very very common mistake. I believe everyone has done it at least once. The important thing is to avoid it in the future :) – xavierm02 Sep 25 '13 at 19:41
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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.

EricAm
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Hint:

Think about the signs of the numerator and denominator in the last step. If $x\to -\infty$, then what happens to $2x + 7$? If you take the positive square root on the denominator, then what is the sign of the quotient?

laughing_man
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