I want to prove that the following sequence $x_n = \frac{3+\sqrt{n}}{2n-\sqrt{n}}$ converges and has a limit.$$\lim_{n \to \infty} \frac{3+\sqrt{n}}{2n-\sqrt{n}} \Rightarrow \lim_{n \to \infty} \frac{\frac{3}{n}+\frac{\sqrt{n}}{n}}{\frac{2n}{n}-\frac{\sqrt{n}}{n}} \Rightarrow \lim_{n \to \infty} \frac{0+0}{2+0}=0$$I want to show that for any $\epsilon$>0 there exists an N such that for all n > N, $|x_n-L|$< $\epsilon$ where L = 0
Let $\epsilon$>0. I want to find N such that for all n>N, $|x_0-0 |$< $\epsilon$$$|x_n-0| =|\frac{3+\sqrt{n}}{2n-\sqrt{n}}-0| = \frac{3+\sqrt{n}}{2n-\sqrt{n}} < \epsilon$$
I now choose N≥max($\frac{6}{\epsilon},\frac{4}{\epsilon^2}$), so for all n>1, then if:$$n≥N \Rightarrow \frac{3+\sqrt{n}}{2n-\sqrt{n}} <\frac{3+\sqrt{n}}{n}= \frac{3}{n}+\frac{1}{\sqrt{n}}< \frac{3}{\frac{6}{\epsilon}} +\frac{1}{\sqrt{\frac{4}{\epsilon^2}}}=\frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon $$
Is this the correct way to do it? I know I made a mistake, because my TA told me that it seemed like a rough draft, but unfortunately didn't have enough time to explain it further.