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

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

• 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? Apr 24, 2012 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. Apr 24, 2012 at 23:49
• Yes, "abstract duplicate" has little meaning for students whose background is mainly computational. Apr 25, 2012 at 0:41
• Jan 30, 2016 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. Jan 30, 2016 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:

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

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 $\frac{1}{n}\sqrt{\square}=\sqrt{\frac{1}{n^2}\square}$.

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.)

• 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 $1/x$.) Apr 24, 2012 at 23:46
• @MichaelJoyce, that's the small technicality I meant. And thanks for finding that typo.
– lhf
Apr 24, 2012 at 23:50
• $x\to 0^+$ is used when you wrote $\sqrt{\frac{1+x}{x^2}}=\frac{\sqrt{1+x}}{x}$ because $\sqrt{x^2}=|x|$. Nov 2, 2017 at 20:04