I missed this question on a test and I can't figure out what I did wrong. I tried asking and was told that I messed up the algebra but then I went home tried it again and got the same answer. The question is $$\text{Find } \lim_{n \to \infty} \frac{2n-2}{n+9}. \text{What is the smallest integer N when } \epsilon = 0.5 $$
Here is my scratchwork: $$\left| \frac{2n-2}{n+9} - \frac{2(n+9)}{n+9} \right| = \left| \frac{2n-2-2n-9}{n+9} \right| = \left| \frac{-11}{n+9} \right| = \frac{11}{n+9}$$
For the proof portion I set $N = \lceil 11/\epsilon \rceil$ (since $\frac{11}{n+9} < \frac{11}{n} < \epsilon $) which was marked correctly but then for the smallest integer I plugged in $\epsilon = 0.5$ into $\frac{11}{\epsilon}-9 = \frac{11}{1/2} - 9 = 22-9 = 13$. So I concluded that the smallest $N = 13$. This is the portion that was marked incorrectly and I'm not sure why aside from that it may have been unclear since I suddenly added the 9 back in? I can't see it so I'm hoping someone here can tell me where my mistake is.