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Let $a_n$ be the $n$-th term of an arithmetic progression with the initial term $a_1=2$ and with the common difference 5. That is, $a_n=2+5(n-1).$ Evaluate $$\lim_{n\rightarrow\infty}n(\sqrt{a_n^2+3}-\sqrt{a_n^2-3}).$$

Well I consider that arithmetic progression here is increasing from 2 to infinity which makes me think that first term in the bracket is greater than the second one. So I think that answer should be positive infinity. But it is not. Can someone help me with it.

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    $\begingroup$ Please use the mathjax latex support and get rid of the image. $\endgroup$ – Aryabhata Jul 14 '17 at 21:19
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$$\sqrt{a_n^2+3}-\sqrt{a_n^2-3} = (\sqrt{a_n^2+3}-\sqrt{a_n^2-3})\cdot\frac{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}}{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}} = \\ = \frac{a_n^2+3 - (a_n^2-3)}{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}}$$

Simplify this expression, and then also use the fact that

$$\sqrt{an^2 + bn + c} =n\cdot\sqrt{1 + \frac bn +\frac c{n^2}}$$

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$$\lim_{n\rightarrow+\infty}n\left(\sqrt{a_n^2+3}-\sqrt{a_n^2-3}\right)=\lim_{n\rightarrow+\infty}\frac{6n}{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}}=$$ $$=\lim_{n\rightarrow+\infty}\frac{6n}{\sqrt{25n^2-30n+12}+\sqrt{25n^2-30n+6}}=\frac{3}{5}$$

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it is $$\frac{6n}{\sqrt{(2+5(n-1))^2+3}+\sqrt{(2+5(n-1))^2-3}}$$ it is equal to $$\frac{6n}{\sqrt{25n^2-30n+12}+\sqrt{25n^2-30n+6}}$$

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