# Calculate $\lim_{n\to{+}\infty}{(\sqrt{n^{2}+n}-n)}$ [duplicate]

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

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## marked as duplicate by Pedro Tamaroff♦, cardinal, Aryabhata, Zev ChonolesApr 25 '12 at 5:21

What have you tried so far? Do you know any algebraic techniques to rewrite expressions that are written as differences where at least one of the terms is a square root? – Michael Joyce Apr 24 '12 at 23:13
@Peter: I don't think that thread does a very good way of avoiding duplicates, as most people who are asking about problems such as the one in this thread are not yet prepared to comprehend the general solution. – Michael Joyce Apr 24 '12 at 23:49
Yes, "abstract duplicate" has little meaning for students whose background is mainly computational. – André Nicolas Apr 25 '12 at 0:41
– Martin Sleziak Jan 30 at 4:19
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 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. – Martin Sleziak Jan 30 at 21:26

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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You can just take $x \rightarrow 0^+$ the whole way through and conclude as you do, since if $\lim_{x \rightarrow 0} g(x)$ exists, then so does $\lim_{x \rightarrow 0^+} g(x)$ and the two limits are equal. (Minor point: you have a $1/n$ that should be a $1/x$.) – Michael Joyce Apr 24 '12 at 23:46
@MichaelJoyce, that's the small technicality I meant. And thanks for finding that typo. – lhf Apr 24 '12 at 23:50