I'm working with limits at infinity and stumbled upon this exercise where I want to evaluate the indicated limit:

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

I tried to solve it by doing the following:

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

But the answer should be $-1$, so what I did must be wrong. How do you evaluate this limit the best way possible?

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    $\begingroup$ For these types of limits, it's usually fruitful to multiply top and bottom by the conjugate of the bottom. $\endgroup$ – David Mitra Jan 26 '14 at 13:44
  • $\begingroup$ Ah, I see. Thanks! $\endgroup$ – dfroli Jan 26 '14 at 13:54
  • $\begingroup$ By the way, what you did isn't "wrong", it just doesn't help you anyhow in this case. $\endgroup$ – Dahn Jan 26 '14 at 13:56
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    $\begingroup$ Be careful when you replace $\sqrt{x^2}$ by $x$. In this case this is allowed because for large $x \gg 0$ we have $\sqrt{x^2} = x$ but in general $\sqrt{x^2} = |x|$, which is equal to $-x$ for $x < 0$. $\endgroup$ – TMM Jan 26 '14 at 14:16
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    $\begingroup$ If you're coming here because you're forgetting to multiply by the conjugate, then something is seriously wrong. This is the "check engine light" to a calculus class. Take heed and study your butt off. Integration is gonna getcha if you don't remember all these little rules!!! $\endgroup$ – alvonellos Jan 26 '14 at 18:11


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The points that we can add to Adi's post are:

$$\frac{1}{\sqrt{x^2-2x}-x}\times\frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x}=\frac{\sqrt{x^2-2x}+x}{(x^2-2x)-x^2}=\color{blue}{\frac{|x|\sqrt{1-2/x}+x}{-2x}}$$ Now if $x\to+\infty$ so $|x|=x$ and so $$\color{blue}{\frac{|x|\sqrt{1-2/x}+x}{-2x}}=\frac{x\sqrt{1-2/x}+x}{-2x}=\frac{x\times \sqrt{1-0}+x}{-2x}\longrightarrow -1$$ And if $x\to-\infty$ so $|x|=-x$ and so $$\color{blue}{\frac{|x|\sqrt{1-2/x}+x}{-2x}}\longrightarrow 0$$

enter image description here

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  • $\begingroup$ Makes sense, thank you! Just one more thing so I'm one hundred percent sure, how do you know that x should be replaced by 1 in the end? $\endgroup$ – dfroli Jan 26 '14 at 14:32
  • $\begingroup$ Cheers Babak, have a nice day! $\endgroup$ – Adi Dani Jan 26 '14 at 14:33
  • $\begingroup$ @AdiDani: The same for you Adi. :-) $\endgroup$ – mrs Jan 26 '14 at 14:36
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    $\begingroup$ @dfroli: I added some steps. $\endgroup$ – mrs Jan 26 '14 at 14:36

What you did is correct but you did not go to the end.

You know that, when $y$ is small, $\sqrt{1-y}$ is close to $1 - y /2$ (this is the begining of the Taylor series); so, the denominator of your last fraction is approximated by : $(1 - 2 / (2 x) - 1) = - 1 / x$.

Dividing the numerator by this last result gives the limit of $-1$ you were expecting.

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  • $\begingroup$ @Adi Dani. Thanks for editing for the old man. Cheers. $\endgroup$ – Claude Leibovici Jan 26 '14 at 14:21
  • $\begingroup$ You are wellcome. $\endgroup$ – Adi Dani Jan 26 '14 at 14:25

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