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I need to find the limit of this sequence:

$a(n)=\sqrt{n^2+9}-\sqrt{n^2-n+9}$

So I multiply with this since $(a-b)(a+b)=(a^2-b^2)$

$\dfrac{\sqrt{n^2+9}+\sqrt{n^2-n+9}}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$

And get $\dfrac{9-n+9}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$

Then divide by n?

And get $\dfrac{9/n-n/n+9/n}{\sqrt{n^2/n+9/n}+\sqrt{n^2/n-n+9/n}}$

$\lim = \dfrac{1}{\infty} = 0$?

Martin Sleziak
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CollapsRanger2
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    First, you multiplied incorrectly...the numerator should be $(n^2+9)-(n^2-n+9)=n$. Secondly, if you mean to divide numerator and denominator by $n$, that's fine, but I don't understand how you got the $\frac 1n$ inside the radical. – lulu Sep 22 '16 at 19:25
  • Hint: in the last step, when you divide to $n $, you have to divide to $n^2$ the quantities under the square roots. – Anatoly Sep 22 '16 at 19:27
  • A suggestion: $a_n$ is easy to compute. Why not evaluate it for modest $n$? Does it look like it is going to $0$? – lulu Sep 22 '16 at 19:27
  • BTW you can find post about rather similar limit of $\sqrt{n^2+n}-n$ here and there are a few other similar questions (and answers). And I'd also like to say that I really appreciate that when posting the question you have shown your own attempt at the solution. Welcome to Math.SE! – Martin Sleziak Sep 24 '16 at 06:32

2 Answers2

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$$\dfrac{n}{(\sqrt{n^2+9}+\sqrt{n^2-n+9})}=\dfrac{n/n}{\frac{1}{n}(\sqrt{n^2+9}+\sqrt{n^2-n+9})}$$ $$=\dfrac{1}{\sqrt{n^2/n^2+9/n^2}+\sqrt{n^2/n^2-n/n^2+9/n^2}}$$

E.H.E
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After multiplying the initial equation by $$\dfrac{\sqrt{n^2+9}+\sqrt{n^2-n+9}}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$$ You should get $(n^2+9) - (n^2 - n +9) = n$ in the numerator, i.e., you should have this:

$$\frac{n}{\sqrt{n^2 + 9}+\sqrt{n^2-n+3}}$$

Now try dividing numerator and denominator by $n$:

$$\lim_{n\to \infty} \frac n{\sqrt{n^2 + 9}+\sqrt{n^2-n+3}}= \lim_{n\to \infty} \frac 1{\sqrt{1+\frac 9{n^2}} +\sqrt{1-\frac 1n+\frac 9{n^2}}} = \frac 12$$

Mikasa
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amWhy
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