By the definition of a limit (using Given an $\epsilon > 0$), prove
$$\lim_{n\to\infty}\frac{6n+2}{3n+5} = 2$$
Proof:
Given epsilon>0 , let N = 1/(3epsilon)
if n >= N, then absolute value of [(6n+2)/(3n+5) -2] = absolute value of [ 12/ (3n+5)] < 1/3n <= 1/3N = epsilon
Therefore $$\lim_{n\to\infty}\frac{6n+2}{3n+5} = 2$$
Is that first proof correct? as well as,
$$\lim_{n\to\infty}\frac{n^2 -2n-3}{2n^2-1} = \frac12$$
n goes to infinity
I'm pretty stuck right now, any help would be great.
Let $\varepsilon>0$, $$\left|\frac{6n+2}{3n+5}-2\right|<\varepsilon\Leftrightarrow n>\frac{8-5\varepsilon}{3\varepsilon}$$ Hence, $\forall\varepsilon>0$, choose $N=max\{\frac{8-5\varepsilon}{3\varepsilon},0\}$. Then $\forall n\geq N$, $\left|\frac{6n+2}{3n+5}-2\right|<\varepsilon$, which ends your proof.