Limit of this sequence: $\lim_{n \to \infty}\left(\sqrt{n^2+2n+5}-n\right)$

Question

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?

Answer 1

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:

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

Answer 2

\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*}

Answer 3

$\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} $

Answer 4

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 $1$.

Answer 5

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 $n+1$.