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I am searching for a good explanation (whether it be analytical or graphical) for why if I have this limit $$\lim_{x\to+\infty} \sqrt{x^2 - 2x + 1} - x$$ I cannot use the hierarchy of infinities like $$\sqrt{x^2\left(1 - \frac{2}{x} + \frac{1}{x^2}\right)} - x \longrightarrow \sqrt{x^2} - x \longrightarrow x - x = 0$$

I know the initial limit shows an indeterminate form $\infty - \infty$, but I also know hierarchy of infinities shall work too. In any case, I know the limit must be computed in such a way to remove the minus sign (and hence the indeterminate form) like

$$\lim_{x\to+\infty} \sqrt{x^2 - 2x + 1} - x \cdot \dfrac{\sqrt{x^2 - 2x + 1} + x}{\sqrt{x^2 - 2x + 1} + x} = \lim_{x\to+\infty} \dfrac{x^2 - 2x + 1 - x^2}{\sqrt{x^2 - 2x + 1} + x}$$

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

The questions hence are:

$\bullet$ Why the initial method leads to a wrong answer?

$\bullet$ Is the "correct" method just a trick used to remove the minus sign, in order to have a sum and hence no indetermination problem?

Thank you!

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The initial method fails because you still have $\infty-\infty$ if you ignore higher order terms.$$\sqrt{x^2\left(1-\frac2x+\frac1{x^2}\right)}=x\sqrt{1-\frac2x+\frac1{x^2}}$$ In the zero-th order approximation the square root is $1$, but in the first order $$\sqrt{1-\alpha}\approx 1-\frac\alpha2+\mathcal{O}(\alpha^{-2})$$ So $$x\sqrt{1-\frac2x+\frac1{x^2}}\approx x\left(1-\frac12\frac2x\right)+x\mathcal{O}(x^{-2})=x-1+\mathcal{O}(x^{-1})$$

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  • $\begingroup$ Thank you so much, that made it perfectly clear! $\endgroup$ Jan 20 at 17:37
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The initial method doesn't work because, when you are dealing with a limit with respect to a variable $x$, you cannot replace part of the expression by its limit and to leave the rest of the expression unchanged. Otherwise, things like this can occur:\begin{align}1&=\lim_{x\to\infty}1\\&=\lim_{x\to\infty}\frac xx\\&=\lim_{x\to\infty}x\cdot\frac1x\\&=\lim_{x\to\infty}x\times0\\&=0.\end{align}

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The other answers already explain why the initial method does not work, so I will only give an easiest way to derive this limit : just notice that for $x > 1$, one has $$\sqrt{x^2 - 2x + 1} - x = \sqrt{(x-1)^2} - x = -1$$

So $$\boxed{\lim_{x\to+\infty} \sqrt{x^2 - 2x + 1} - x = -1}$$

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  • $\begingroup$ The completion of the square! Or, in this case, the exact square. Nice easy explanation (which clearly holds in general) :) $\endgroup$ Jan 20 at 17:38

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