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So guys I need to find the limit of:

$\displaystyle\lim_{n \to \infty}\left(\sqrt{n^2+2n+5}-n\right)$

The quadratic equation is hard to factorise and I really struggle to answer these questions.

Do I need to use the limit comparison test after I simplify this?

Martin Sleziak
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MooeSha123
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5 Answers5

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When you're facing a limit that's difference of square roots (or any radical indeed) it's a nice idea is to multiply by the sum of the two square roots. So we have:

$$\lim_{n \to \infty} \sqrt{n^2 + 2n + 5} - n = \lim_{n \to \infty} \frac{n^2 + 2n + 5 - n^2}{\sqrt{n^2+2n+5}+n} = \lim_{n \to \infty} \frac{2 + \frac 5n}{\sqrt{1 + \frac 2n + \frac 5{n^2}} + 1} = 1$$

Stefan4024
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  • Should i delete my answer ? just posted a 1 minute after you and found out that it is a duplicate. –  Jun 26 '16 at 16:52
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    @ritwiksinha I leave it up to you. I mean I don't own any copyrights to this answer after all. ;) – Stefan4024 Jun 26 '16 at 16:53
  • I will wait till i receive a down vote. :) –  Jun 26 '16 at 16:55
  • Thanks that helped alot, now suppose I had to find the limit of the sum of the sequence, what would I have to do differently? – MooeSha123 Jun 26 '16 at 17:26
  • @MooeSha123 Well, if $\sum_{n=0}^{\infty} \left(\sqrt{n^2+2n+5} - n\right)$ converges then $\lim_{n \to \infty} \left(\sqrt{n^2+2n+5} - n\right) = 0$. But as this is not true the series is divergent. In fact $\sum_{n=0}^{\infty} \left(\sqrt{n^2+2n+5} - n\right) \approx \lim_{n \to \infty} n = \infty$ – Stefan4024 Jun 26 '16 at 17:30
  • If you put in n=1 rather than where you put n=0, would that make a difference? – MooeSha123 Jun 26 '16 at 17:33
  • @MooeSha123 Not at all. If we add or subtract finately many finite terms to the sequence that wouldn't change the convergence/divergence of the series. – Stefan4024 Jun 26 '16 at 17:42
  • And how about the limit for the summation sequence? – MooeSha123 Jun 26 '16 at 17:55
  • @MooeSha123 Summation sequence? I don't get the question – Stefan4024 Jun 26 '16 at 18:04
  • The limit of the sequence itself. Is is infinity? – MooeSha123 Jun 26 '16 at 18:06
  • @MooeSha123 What sequence? We proved that the limit of the sequence in your question is $1$. – Stefan4024 Jun 26 '16 at 18:07
  • No I mean the sum sequence as in what you were referring to in your last line when I asked for the limit of the sum of sequence. – MooeSha123 Jun 26 '16 at 18:13
  • @MooeSha123 I recommend you to close this question by selecting an answer and what you asked is already answered. –  Jun 26 '16 at 18:13
  • @MooeSha123 if $a_{n + 1} - a_{n} \lt 1$ only then the series will converge, else not. –  Jun 26 '16 at 18:35
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\begin{align*} \lim_{n \to \infty} (\sqrt{n^2 + 2n +5} - n) \cdot \frac{(\sqrt{n^2 + 2n +5} + n)}{(\sqrt{n^2 + 2n +5} + n)} &=\lim_{n \to \infty} \frac{2n +5}{(\sqrt{n^2 + 2n +5} + n)}\\ &=\lim_{n \to \infty} \frac{2 +{5\over n}}{(\sqrt{1 + {2\over n} +{5\over n^2}} + 1)}\\ &=\lim_{n \to \infty} \frac{2}{(1 + 1)} = 1 \end{align*}

clark
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$\begin{array}\\ \sqrt{n^2+2n+5}-n &=n(\sqrt{1+2/n+5/n^2}-1)\\ &= n((1+(2/n+5/n^2)/2+O(1/n^2))-1)\\ &= n(1/n+5/(2n^2)+O(1/n^2))\\ &= 1+5/(2n)+O(1/n)\\ &\to 1 \qquad\text{as } n \to \infty\\ \end{array} $

marty cohen
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Multiplying top and bottom by $\sqrt{n^2+2n+5}+n$ is almost automatic here, and is the "right" way to proceed. But let us take another approach. Note that $$n^2+2n+5=(n+1)^2+4\gt (n+1)^2.$$ Also, $$(n+1)^2+4\lt \left(n+1+\frac{2}{n+1}\right)^2.$$ This inequality is easy to verify by squaring $n+1+\frac{2}{n+1}$. Thus $$1\lt \sqrt{n^2+2n+5}-n\lt 1+\frac{2}{n+1}.$$ So by Squeezing our limit is $1$.

Remark: The above is a formal version of the observation that for $n$ large, $\sqrt{(n+1)^2+4}$ is "almost" equal to $n+1$.

André Nicolas
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A round-about but fun/elementary approach. Note that \begin{align*} \lim_{n\rightarrow\infty}\sqrt{n^{2}+2n+5}-n & =\lim_{n\rightarrow\infty}\sqrt{n\left(n+2\right)+5}-n\\ & =\lim_{m\rightarrow\infty}\sqrt{\left(m-1\right)\left(m+1\right)+5}-\left(m-1\right)\\ & =\lim_{m\rightarrow\infty}\sqrt{m^{2}+4}-m+1 \end{align*} so that it is enough to consider $$ \lim_{m\rightarrow\infty}\sqrt{m^{2}+4}-m. $$ Clearly, for $m\geq0$, the expression under the limit is positive. Let $\epsilon>0$, and note that \begin{align*} \sqrt{m^{2}+4}-m<\epsilon & \iff\sqrt{m^{2}+4}<\epsilon+m\\ & \iff m^{2}+4<\left(\epsilon+m\right)^{2}\\ & \iff m^{2}+4<\epsilon^{2}+2\epsilon m+m^{2}\\ & \iff0<\epsilon^{2}+2\epsilon m-4. \end{align*} Clearly, we can pick $m$ large enough so that this is true, and hence $$ \lim_{m\rightarrow\infty}\sqrt{m^{2}+4}-m=0 $$ so that the original limit is $1$.

parsiad
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