Prove using $\epsilon - n_0$:
$$\lim_{n\rightarrow\infty}\frac{n^2+2n-2}{n^2-6}=1$$ I made an attempt on this question but I'm not sure if this is the correct path.
Firstly I did $\frac{n^2+2n-2}{n^2-6}-1$ < $\epsilon$ ?
after that |$\frac{n^2+2n-2}{n^2-6}-1$| = |$\frac{2(n+2)}{n^2-6}$| which =|$\frac{2n+4}{n^2-6}$|.
What do I do after this?
Would $n\geq3$ due to making the denominator positive?
$$n\geq 3 \implies 2n\geq6 \implies 2n+4 \geq 10 > 0$$
Are these steps correct?
Notice that $(2n+4)(n-\frac32)=2n^2+n-6>n^2-6$
Then for $n>3/2$ we have that
$$\left|\frac{2(n+2)}{n^2-6}\right|< \frac{2n+4}{(2n+4)(n-3/2)}=\frac1{n-3/2}<\epsilon$$ for an $n>3/2$ large enough.