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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:

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

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

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    Please define $[x]$. Do you mean the integer part, a.k.a. the "floor function", $\lfloor x \rfloor$? – 2017-01-13
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    @FlybyNight [x]="floor function – 2017-01-13
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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 $n$? – 2017-01-13
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    @Arthur You should make this into an answer. – 2017-01-13
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    The limit of this function is within 0 to 1. – 2017-01-13
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    @Arthur it depends on the n,so this function doesn't have limits. What I mean is theat the limit is indeterminate – 2017-01-13
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    @Namasivayam Kalithasan How do you know that the function doesn't have a limit? – 2017-01-13
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    @Arthur Come on. Stop being a tease and post your answer! – 2017-01-13
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    @FlybyNight there is already an answer which quantifies that difference (and is almost correct in doing so). – 2017-01-13
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    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? – 2017-01-13
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    @FlybyNight Your answer is more than good enough, if only you correct the minor flaw and undelete it. – 2017-01-13
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    Quite easy...its obvious that x and floor (x) is different at most by 1@fernando – 2017-01-13
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    @NamasivayamKalithasan So, you conclude "quite eas[il]y" that the limit does not exist, while it *does* exist? – 2017-01-13
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    @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 – 2017-01-13
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    @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. – 2017-01-13

2 Answers 2

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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

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

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    How does that answer the question? – 2017-01-13
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    @ClementC. The question asks for the limit and this gives it? – 2017-01-13
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    Oops. Mea culpa. – 2017-01-13
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    @Sil Thanks. Fixed it. – 2017-01-13
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    Did you mean the both N are equal? – 2017-01-13
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    -1 You mean floor $(a_n)$ is n,and this n is integer but in the limit the n is any real number.you're wrong – 2017-01-13
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    @NamasivayamKalithasan If two things are not equal, I don't use the same variable for it. And, no, $n$ is the index of a sequence, so it's not "any reall number". It's a positive integer – 2017-01-13
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    @NamasivayamKalithasan Your comment seems misguided. We are asked for the limit of $f(a_n)$, when $n\to\infty$ (so implicitly $n$ is an integer, as it is the index of the sequence). – 2017-01-13
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Hint: $\sqrt{n^2+2n } = \sqrt{(n+1)^2-1} \approx n+1 - \frac 12 \frac{1}{n+1}$.

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    While the conclusion wouldn't change, your development is wrong. There is no square in the denominator: $$\sqrt{n^2+2n} = n+1-\frac{1}{2n} + o\!\left(\frac{1}{n}\right)$$ – 2017-01-13
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    So $\lim_{ n \to \infty }a_n=\sqrt{n^2+2n}=1?$ – 2017-01-13
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    No. The limit you are asking **in your question** is $1$, the one in your comment is of course $\infty$. (Or, actually, the way written just above, does not even "parse") – 2017-01-13
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    @ClementC. thanks for the catch! – 2017-01-13