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}} $
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}} $
$ \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 $