# How to find a limit of this sequence?

Could you explain it step by step?

$$\lim_{n \to \infty} \frac{\sqrt{n^2+6n+2}-\sqrt{n^2+2}}{\sqrt{9n^2+n}-\sqrt{9n^2-n}}$$

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What have you tried? –  icurays1 Nov 12 '12 at 19:14
I tried to multiply this sequence by $$\frac{\sqrt{9n^2+n}+\sqrt{9n^2-n}}{\sqrt{9n^2+n}+\sqrt{9n^2-n}}$$ –  TomDavies92 Nov 12 '12 at 19:17
That's a good start, since that will 'rationalize' the denominator (the denominator will be $2n$ after the dust clears). You'll have a mess on the top, but you should be able to expand it all out, then factor out a $2n$ to cancel the denominator, and see what's left when $n\rightarrow\infty$. –  icurays1 Nov 12 '12 at 19:25

$$\lim_{n \to \infty} \frac{\sqrt{n^2+6n+2}-\sqrt{n^2+2}}{\sqrt{9n^2+n}-\sqrt{9n^2-n}}\\ =\lim_{n \to \infty} \left(\frac{\sqrt{n^2+6n+2}-\sqrt{n^2+2}}{\sqrt{9n^2+n}-\sqrt{9n^2-n}} \times \frac{\sqrt{9n^2+n}+\sqrt{9n^2-n}}{\sqrt{9n^2+n}+\sqrt{9n^2-n}} \times \frac{\sqrt{n^2+6n+2}+\sqrt{n^2+2}}{\sqrt{n^2+6n+2}+\sqrt{n^2+2}} \right)\\ = \lim_{n \to \infty}\frac{6n}{2n} \frac{\sqrt{9n^2+n}+\sqrt{9n^2-n}}{\sqrt{n^2+6n+2}+\sqrt{n^2+2}}\\ = \lim_{n \to \infty}3 \frac{n\left(\sqrt{9+\frac{1}{n}}+\sqrt{9-\frac{1}{n}}\right)}{n\left(\sqrt{1+\frac{6}{n}+\frac{2}{n^2}}+\sqrt{1+\frac{2}{n^2}}\right)} \\ =3\times \frac{6}{2}\\ =9$$
• Put $n = \infty$. We get the form $\frac{\infty}{\infty}$. Lame! We need to do something else. We know that $\frac{n}{\infty} = 0$. How can I use this fact?
• Divide each term by $n^2$ (why not with $n^3$?).
• Expression becomes $\frac{\sqrt{1+\frac{6}{n}+\frac{2}{n^2}}-\sqrt{1+\frac{2}{n^2}}}{\sqrt{9+\frac{1}{n}}-\sqrt{9-\frac{1}{n}}}$.
• Let's put $n \rightarrow \infty$. We get $\frac{1-1}{3-3}$ i.e. $0/0$.
• I need to do something else. What if I somehow remove 0 from denominator? What can I do when I see an expression of form $a-b$ in denominator. Wait! I can multiply it with $a+b$. So multiply both numerator and denominator with a suitable term and repeat the process. Hopefully, there won't be any zero in the denominator.