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I am working on the following problem.

Find the limit of $$\lim_{n \to \infty} (\sqrt{n^2-n}-n)$$

Intuitively, I want to say it's $0$ because as $n \to \infty$, $\sqrt{n^2-n}$ behaves like $n$ and subtracting $n$ makes it $0$.

However, algebraically, to my surprise

$$\begin{align} \lim_{n \to \infty} (\sqrt{n^2-n}-n) & = \lim_{n \to \infty} \frac{n}{\sqrt{n^2-n}+n}\\ & = \lim_{n \to \infty} \frac{1}{\sqrt{1-\frac{1}{n}}+1}\\ & = \frac{1}{2} \\ \end{align}$$

Is there any intuitive way to explain why 0 was not the answer ?

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Because infinity minus infinity is not always zero. What if the linear term inside the square root was -2n instead of -n? According to your explanation, the answer is still zero? – imranfat May 31 '13 at 22:45
Side note here; the linear term does not influence the limit in this problem, but in other problems, it may. Bottom line: Infinity minus infinity could be anything – imranfat May 31 '13 at 22:47
up vote 15 down vote accepted

You have a sign error: you should have $-\frac12$.

Complete the square: $n^2-n=\left(n-\frac12\right)^2-\frac14$, so for large $n$ its square root is very close to $n-\frac12$, and subtracting $n$ brings you very close to $-\frac12$.

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Thanks, that was a perfect explanation. – hyg17 May 31 '13 at 22:47
@hyg17: You’re welcome. – Brian M. Scott May 31 '13 at 22:49
Topology, perfect sets, uncountability. But I don't want to be dragging you around, do as you please, please. And Baire's Theorem! – Pedro Tamaroff May 31 '13 at 22:51
That was very good explanation Mr. Scott – mathemagician Jun 1 '13 at 6:22

Using $(1+x)^\alpha \approx_0 1+\alpha x$ we have: $$\sqrt{n^2-n}-n=n\left(\sqrt{1-\frac{1}{n}}-1\right)\approx_\infty n\left(1-\frac{1}{2n}-1\right)=-\frac{1}{2}$$

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That's an interesting way of thinking. Thanks. – hyg17 May 31 '13 at 23:04
@hyg17 You're welcome. – user63181 Jun 1 '13 at 7:42

You are wrong, because the expression is clearly negative for $n \gt 0$

One way of seeing the result is to note that $n^2-n=\left(n-\frac 12\right)^2-\frac 14$. As $n$ increases, the $\frac 14$ becomes insignificant.

You should get $-\frac 12$

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Thank you. Apparently it's common sense. – hyg17 May 31 '13 at 22:49

If you know that $\sqrt{1+x}\sim1+\frac x2$ for $x\sim0$, we get that $$ \begin{align} \lim_{n\to\infty}\left(\sqrt{n^2-n}-n\right) &=\lim_{n\to\infty}n\left(\sqrt{1-\frac1n}-1\right)\\ &=\lim_{n\to\infty}n\left(\left(1-\frac1{2n}\right)-1\right)\\ &=\lim_{n\to\infty}n\left(-\frac1{2n}\right)\\ &=-\frac12 \end{align} $$ or to complete your answer and correct the sign: $$ \begin{align} \lim_{n\to\infty}\left(\sqrt{n^2-n}-n\right) &=\lim_{n\to\infty}\left(\sqrt{n^2-n}-n\right)\frac{\sqrt{n^2-n}+n}{\sqrt{n^2-n}+n}\\ &=\lim_{n\to\infty}\frac{(n^2-n)-n^2}{\sqrt{n^2-n}+n}\\ &=\lim_{n\to\infty}\frac{-n}{\sqrt{n^2-n}+n}\\ &=\lim_{n\to\infty}\frac{-1}{\sqrt{1-1/n}+1}\\ &=-\frac12 \end{align} $$

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