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Calculate $\displaystyle\lim_{n \to \infty} \left(\,{\sqrt{\,{n^{2} + n}\,} - n}\,\right)$.

$\displaystyle\lim_{n \to \infty}\left(\,{\sqrt{\,{n^{2} + n}\,} - n}\,\right) = \infty - \infty$ We have an indeterminate form

So I proceeded to factorize $$\sqrt{n^2+n} -n = \sqrt{ \frac{n^2(n+1)}{n}}-n =n \left[ \sqrt{\frac{n+1}{n}}-1 \right]$$

taking the limit: $$\lim\limits_{n \rightarrow \infty} n \left[ \sqrt{\frac{n+1}{n}}-1 \right]= \infty \cdot 0$$

indeterminate again

What am i missing? How is the way forward to proceed? Much appreciated

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marked as duplicate by Hans Lundmark, G-man, user91500, Glorfindel, Michael Hoppe Aug 5 '17 at 10:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Do you know what the derivative of $x\mapsto \sqrt{1+x}$ is at $x=0 ?$ $\endgroup$ – kimchi lover Aug 5 '17 at 2:52
  • $\begingroup$ HINT: $\infty$ is not a number. $\endgroup$ – George N. Missailidis Aug 5 '17 at 2:52
  • $\begingroup$ $n^2$ will ultimately dominate $n$ by any amount. $\endgroup$ – gary Aug 5 '17 at 2:54
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    $\begingroup$ This has been asked countless times before; see for example math.stackexchange.com/questions/136495/… (in particular Martin Sleziak's comment and the lists of linked and related questions). $\endgroup$ – Hans Lundmark Aug 5 '17 at 7:03
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Hint: use the so-to-speak "multiply and divide by the conjugate" trick — it often helps to rationalize. In this case, since you're given a difference $\sqrt{n^2+n}-n$, multiply and divide by the sum of the same two terms $\sqrt{n^2+n}+n$:

$$\lim_{n\to\infty} \left(\sqrt{n^2+n}-n\right)=\lim_{n\to\infty} \frac{\left(\sqrt{n^2+n}-n\right)\left(\sqrt{n^2+n}+n\right)}{\sqrt{n^2+n}+n}=\cdots$$

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hint

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

$$=\frac {1}{\sqrt {1+\frac {1}{n}}+1} $$

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Trick is to write $\sqrt{n^2 + n} - n$ in a nice form.Below we use rationalization

$\lim _{n\rightarrow \infty}\sqrt{n^2 + n} - n = \lim_{n \rightarrow \infty} \sqrt{n}(\sqrt{n+1} - \sqrt{n}) = \lim_{n \rightarrow \infty} \sqrt{n}(\frac{1}{\sqrt{n+1}+\sqrt{n}}) = \lim_{n \rightarrow \infty} \frac{\sqrt{n}(1)}{\sqrt{n}(\sqrt{1+\frac{1}{n}} + 1)} = \frac{1}{2}$

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Hint:

$n^2+n=n(n+1),$ set $n=\cot^2t$

Alternatively,

Use Taylor's expansion for $$\left(1+\dfrac1n\right)^{1/2}$$

Or set $1/n=h$

$$\sqrt{n^2+n}-n=\dfrac{\sqrt{1+h}-1}h=\dfrac{(1+h)-1}{h(\sqrt{1+h}+1)}=?$$

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Using you approach $$\sqrt{n^2+n} -n = \sqrt{ \frac{n^2(n+1)}{n}}-n =n \left[ \sqrt{\frac{n+1}{n}}-1 \right]=n \left[ \sqrt{1+\frac{1}{n}}-1 \right]$$ Now, use the generalized binomial theorem or Taylor series to get, for small values of $x$, $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$$ $$\sqrt{1+x}-1=\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$$ Make now, since $n$ is large, $x=\frac 1n$ $$\sqrt{1+\frac{1}{n}}-1=\frac{1}{2 n}-\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ $$n \left[ \sqrt{1+\frac{1}{n}}-1 \right]=\frac{1}{2}-\frac{1}{8 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.

Try with $n=10$ (which is very small, so you will have $\sqrt{110}-10\approx 0.488088$ while the above truncated series would give $\frac{39}{80}=0.487500$.

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Alternatively: $$L=\lim_\limits{n\to\infty} \left(\sqrt{\left(n+\frac12 \right)^2-\frac14}-n\right)=\lim_\limits{n\to\infty} \left(n+\frac12-n\right)=\frac12.$$

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