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How to determine the convergence behavior of the sequence $x_{n}=\sqrt{n^{2}+11n+21}-\sqrt{n^{2}+6}$ and its limit?

My effort: I don't know where to begin.

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Hint: multiply and divide $x_n$ by $\sqrt{n^2+11n+21}+\sqrt{n^2+6}$

$\sqrt{n^2+11n+21}-\sqrt{n^2+6}= (\sqrt{n^2+11n+21}-\sqrt{n^2+6})({\sqrt{n^2+11n+21}+\sqrt{n^2+6}\over {\sqrt{n^2+11n+21}+\sqrt{n^2+6}}}$

$={{11n+15}\over{\sqrt{n^2+11n+21}+\sqrt{n^2+6}}}$. So the limit is $11/2$.

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Hint: Write $\sqrt{n^2+an+b}=n\sqrt{1+\frac an+\frac b{n^2}}$ and use Taylor expansion $\sqrt{1+x}=1+\frac12 x+O\left(x^2\right)$ as $x\to 0$.

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As G. Sassatelli and Zack Ni answered, Taylor series are very useful not only to get the limit but also how it is approached.

Using, for small $y$, $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ then $$\sqrt{n^2+a n +b}=n \sqrt{1+\frac a n+\frac b {n^2}}=n\left(1+\frac{a}{2 n}+\frac{4b-{a^2}}{8n^2}+O\left(\frac{1}{n^3}\right)\right)$$ $$\sqrt{n^2+a n +b}=n+\frac{a}{2}+\frac{4b-{a^2}}{8n}+O\left(\frac{1}{n^2}\right)$$ So, using your numbers $$x_n=\frac{11}{2}-\frac{61}{8 n}+O\left(\frac{1}{n^2}\right)$$ For illustration purposes $$x_{10}=\sqrt{231}-\sqrt{106}\approx 4.90305$$ while the approximation gives $\frac{379}{80}\approx 4.73750$.

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