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Possible Duplicate:
Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$

Could someone help me through this problem? Calculate $\displaystyle\lim_{n \to{+}\infty}{(\sqrt{n^{2}+n}-n)}$

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    I agree with the above comments that it is not very good idea to close this as a duplicate of a much more general question http://math.stackexchange.com/q/30040. I am voting to reopen. (After the question is reopened, it can be closed as a duplicate of some questions which *really are* duplicates of this one.) See also relevant discussion [in chat](http://chat.stackexchange.com/transcript/2165/2016/1/30).2016-01-30

3 Answers 3

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We have:

$\sqrt{n^{2}+n}-n=\frac{(\sqrt{n^{2}+n}-n)(\sqrt{n^{2}+n}+n)}{\sqrt{n^{2}+n}+n}=\frac{n}{\sqrt{n^{2}+n}+n}$ Therefore:

$\sqrt{n^{2}+n}-n=\frac{1}{\sqrt{1+\frac{1}{n}}+1}$

And since: $\lim\limits_{n\to +\infty}\frac{1}{n}=0$

It follows that:

$\boxed{\,\,\lim\limits_{n\to +\infty}(\sqrt{n^{2}+n}-n)=\dfrac{1}{2}\,\,}$

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Guide: Rationalize,

$\left(\sqrt{n^2+n}-\sqrt{n^2}\right)\cdot \frac{\sqrt{n^2+n}+\sqrt{n^2}}{\sqrt{n^2+n}+\sqrt{n^2}}=\frac{n}{\sqrt{n^2+n}+\sqrt{n^2}}$

Now divide numerator and denominator by $n$. Remember $\frac{1}{n}\sqrt{\square}=\sqrt{\frac{1}{n^2}\square}$.

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Here's an answer that is probably not within the intended scope but it's nice anyway...

Let $x=1/n$. Then $ \lim_{n\to{+}\infty}{\sqrt{n^{2}+n}-n} = \lim_{x\to0}{\sqrt{\frac1{x^2}+\frac1x}-\frac1x} = \lim_{x\to0}{\sqrt{\frac{1+x}{x^2}}-\frac1x} = \lim_{x\to0}{\frac{\sqrt{1+x}}{x}-\frac1x}= \lim_{x\to0}{\frac{\sqrt{1+x}-1}{x-0}} = f'(0) = \frac12 $ for $f(x)=\sqrt{1+x}$.

(There's a small technicality that actually $x\to0^+$ but let's overlook that.)

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    $x\to 0^+$ is used when you wrote $\sqrt{\frac{1+x}{x^2}}=\frac{\sqrt{1+x}}{x}$ because $\sqrt{x^2}=|x|$.2017-11-02