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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:
$\sqrt{n^2 + n} - n \lt \sqrt{n^2 +n + \frac{1}{4}} - n = n+\frac{1}{2} - n = 1/2$

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  • $\begingroup$ You can try a series expansion. $\endgroup$
    – Amzoti
    Feb 19, 2013 at 0:06
  • $\begingroup$ Just write $\sqrt{n^2 + n} = n \sqrt{1 + \frac{1}{n}} \rightarrow n$ $\endgroup$
    – vonbrand
    Feb 19, 2013 at 1:12
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    $\begingroup$ @vonbrand: the correct answer is $\frac12$, not $0$. Just because $\sqrt{1+\frac{1}{n}} \to 1$, it doesn't follow that $n\sqrt{1+\frac{1}{n}} - n \to 0$. $\endgroup$
    – TonyK
    Jan 4, 2016 at 22:37
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    $\begingroup$ Possible duplicate of Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ $\endgroup$
    – Aryabhata
    Jan 4, 2016 at 22:50
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    $\begingroup$ @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. $\endgroup$
    – user147263
    Jan 4, 2016 at 23:44

3 Answers 3

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

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    $\begingroup$ Can't believe I forgot about that trick! $\endgroup$
    – Mark
    Feb 19, 2013 at 0:14
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    $\begingroup$ @Mark: When in doubt, multiply by $1$, or add $0$. :) $\endgroup$ Feb 19, 2013 at 0:15
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$$\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n =\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n\frac{\sqrt{n^2 + n}+n}{\sqrt{n^2 + n}+n}=...=\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^2 + n}+n}=...=\frac{1}{2}$$

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    $\begingroup$ Just noting that this is the same solution as @gnometorule for anyone who is skimming over the solutions. $\endgroup$
    – Mark
    Sep 11, 2014 at 6:15
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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.

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  • $\begingroup$ I think this is essentially the strategy suggested by @Amzoti with expansion being just the first order taylor approximation = MVT. $\endgroup$
    – Mark
    Sep 11, 2014 at 6:11

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