Computing limits which involve square roots, such as $\sqrt{n^2+n}-n$

Question

Is there a general strategy for this? For example I'm working on the limit
$$\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n $$

I have a simple argument to show that this limit is less than or equal to 1/2, but I can't get much further because it's difficult for me to manipulate the square root symbol. Here is the argument:
$\sqrt{n^2 + n} - n \lt \sqrt{n^2 +n + \frac{1}{4}} - n = n+\frac{1}{2} - n = 1/2$

Just write $\sqrt{n^2 + n} = n \sqrt{1 + \frac{1}{n}} \rightarrow n$ — vonbrand, Feb 19 '13 at 01:12
@vonbrand: the correct answer is $\frac12$, not $0$. Just because $\sqrt{1+\frac{1}{n}} \to 1$, it doesn't follow that $n\sqrt{1+\frac{1}{n}} - n \to 0$. — TonyK, Jan 04 '16 at 22:37
Possible duplicate of Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ — Aryabhata, Jan 04 '16 at 22:50
Please consider browsing the [faq] tag before posting a question. [To anyone who comes across this comment, I know this question is old] — Aryabhata, Jan 04 '16 at 22:50
@Aryabhata The ubiquitous multiply-by-conjugate trick is clearly explained here, and not so much in the linked post, because of the greater generality of the latter. A special case is not always a duplicate. — , Jan 04 '16 at 23:44
@Normal: I suggest you read this: http://meta.math.stackexchange.com/questions/1868/list-of-generalizations-of-common-questions and the linked question by Bill D. Special case vs general case is a common objection to such closings, but such closings [i.e. closed as minor variations] are fine IMO. It does not have to be an exact dupe. — Aryabhata, Jan 06 '16 at 00:10

score 15 · Accepted Answer · edited Nov 19 '17 at 13:48

15

$\begin{align} \sqrt{n^2 + n} - n & = (\sqrt{n^2 + n} - n) \cdot \frac{\sqrt{n^2 + n} + n}{\sqrt{n^2 + n} +n} \\ & = \frac{n^2+n-n^2}{\sqrt{n^2 + n} + n} \\ & = \frac{1}{\frac{\sqrt{n^2 + n} + n}{n}} \\ & = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} \rightarrow \frac{1}{2}\\ \end{align}$

edited Nov 19 '17 at 13:48

Trevor Gunn

27,041

answered Feb 19 '13 at 00:12

gnometorule

4,640

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Can't believe I forgot about that trick! – Mark Feb 19 '13 at 00:14
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@Mark: When in doubt, multiply by $1$, or add $0$. :) – gnometorule Feb 19 '13 at 00:15

score 6 · Answer 2 · answered Feb 19 '13 at 00:10

6

$$\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n =\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n\frac{\sqrt{n^2 + n}+n}{\sqrt{n^2 + n}+n}=...=\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^2 + n}+n}=...=\frac{1}{2}$$

answered Feb 19 '13 at 00:10

Adi Dani

16,949

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Just noting that this is the same solution as @gnometorule for anyone who is skimming over the solutions. – Mark Sep 11 '14 at 06:15

score 1 · Answer 3 · answered Sep 11 '14 at 06:10

By the bound stated in the problem, it suffices to show that the limit is $\ge \frac{1}{2}$. This proof is a lot worse than the others but maybe a good exercise in mean value theorem which is why I'm writing it up. Intuitively we want a way to estimate how far $\sqrt{n^2 + n}$ is from $n$. But the mean value theorem along with concavity gives exactly the approximation we need.

Let $f_n(x) = \sqrt{n^2 + x}$ $$f_n'(x) = \frac{1}{2}(n^2 + x)^{-\frac{1}{2}}$$ Now use the mean value theorem $$\sqrt{n^2 + n} - n = f_n(n) - f_n(0) = f_n'(x^*)n$$ where $x^* \in (0, n)$ (and I have suppressed dependence of $x^*$ on $n$). Then by the fact that $f_n'$ is decreasing, we get $$ f_n'(n) \lt f_n'(x*)$$ Multiplying by $n$ on both sides $$ n f_n'(n) < n f_n'(x^*) = \sqrt{n^2 + n} - n$$ Substituting $$ \frac{1}{2}\frac{n}{(n^2 + n)^{\frac{1}{2}}} \lt \sqrt{n^2 + n} - n$$ $$ \frac{1}{2}\frac{1}{{(1 + 1/n)}^{\frac{1}{2}}} \lt \sqrt{n^2 + n} - n$$

And actually the upper bound of $1/2$ could have been found using the same method.

I think this is essentially the strategy suggested by @Amzoti with expansion being just the first order taylor approximation = MVT. — Mark, Sep 11 '14 at 06:11

Computing limits which involve square roots, such as $\sqrt{n^2+n}-n$

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