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If $a_n=\sqrt{n^2+2n}$ and $f(x)=x-\lfloor x \rfloor$, where $\lfloor x \rfloor$ is the floor function, then what is the limit $$\lim_{ n \to \infty }f(a_n) \ \ ?$$

I tried:

$\lim_{ n \to \infty }a_n=\lim_{ n \to \infty }\sqrt{n^2+2n}=\infty$

$\lim_{ n \to \infty }f(\infty)=?$

egreg
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Almot1960
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  • Please define $[x]$. Do you mean the integer part, a.k.a. the "floor function", $\lfloor x \rfloor$? – Fly by Night Jan 13 '17 at 17:40
  • @FlybyNight [x]="floor function – Almot1960 Jan 13 '17 at 17:42
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    Note that because of the floor function, $f(x)$ doesn't care how large $x$ is, only how far away it is from being an integer. Therefore, it's not relevant at all that $\lim_{n \to \infty}a_n = \infty$. You should rather turn your focus on this: How far is $\sqrt{n^2 + 2n}$ from being an integer, for very large $n$? – Arthur Jan 13 '17 at 17:42
  • @Arthur You should make this into an answer. – Fly by Night Jan 13 '17 at 17:44
  • The limit of this function is within 0 to 1. –  Jan 13 '17 at 17:45
  • @Arthur it depends on the n,so this function doesn't have limits. What I mean is theat the limit is indeterminate –  Jan 13 '17 at 17:47
  • @Namasivayam Kalithasan How do you know that the function doesn't have a limit? – Fernando Jan 13 '17 at 17:49
  • @Arthur Come on. Stop being a tease and post your answer! – Fly by Night Jan 13 '17 at 17:58
  • @FlybyNight there is already an answer which quantifies that difference (and is almost correct in doing so). – Clement C. Jan 13 '17 at 17:59
  • Also, am I the only one bothered by the fact that writing $$\lim_{ n \to \infty }a_n=\sqrt{n^2+2n}=\infty$$ is barely a valid notation? – Clement C. Jan 13 '17 at 18:01
  • @FlybyNight Your answer is more than good enough, if only you correct the minor flaw and undelete it. – Arthur Jan 13 '17 at 18:01
  • Quite easy...its obvious that x and floor (x) is different at most by 1@fernando –  Jan 13 '17 at 18:02
  • @NamasivayamKalithasan So, you conclude "quite eas[il]y" that the limit does not exist, while it does exist? – Clement C. Jan 13 '17 at 18:04
  • @Clement c , I'm a mathematician ,I won't conclude just by seeing that fact but it too play a very crucial role in that limit –  Jan 13 '17 at 18:11
  • @NamasivayamKalithasan I guess I was not clear. (1) You claim that it is easy to see that the limit does not exist. Yet, (2) the limit does exist (it is equal to $1$), as one of the answers shows, and one can see for instance by an asymptotic development of $a_n$. Hence my comment, to point out that (1) and (2) are, let's say, hard to reconcile. – Clement C. Jan 13 '17 at 18:13

2 Answers2

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We have $n \leq \sqrt{n^2} < \sqrt{n^2+2n} <\sqrt{(n+1)^2} =n+1$, so $\lfloor a_n \rfloor = n$.

Then

$$\lim_{n \to \infty} f(a_n) = \lim_{n \to \infty}\sqrt{n^2+2n} - n =\lim_{n \to \infty} \frac{(\sqrt{n^2+2n} - n)(\sqrt{n^2+2n} + n)}{\sqrt{n^2+2n} + n} $$ $$= \lim_{n\to \infty} \frac{2n}{\sqrt{n^2+2n} + n} = 1.$$

B. Goddard
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Hint: $\sqrt{n^2+2n } = \sqrt{(n+1)^2-1} \approx n+1 - \frac 12 \frac{1}{n+1}$.

TZakrevskiy
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