Prove the limit using definition of limit$$ \ \lim_{a \to \infty}\ \frac{3a+2}{5a+4}= \frac{3}{5} $$
Answer: Let $\varepsilon \ >0 $. We want to obtain the inequality $$\left|\frac{3a+2}{5a+4}- \frac{3}{5}\right|< {\varepsilon}$$
$$ \Rightarrow \left|\frac{3a+2}{5a+4} - \frac{3}{5}\right|\ =\left|\frac{5(3a+2)-3(5a+4)}{5(5a+4)}\right|\\= \left|\frac{-2}{5(5a+4)}\right|\le\frac{1}{a} $$
Therefore we choose $K \in N$ s.t $K> \frac{1}{\varepsilon} $
$$\Rightarrow\left|\frac{3a+2}{5a+4}- \frac{3}{5}\right| \le\frac{1}{a}\le\frac{1}{K} < \varepsilon $$ Is this correct?
let $ζ>0$ be any arbitrary real number consider, $$\begin{align} &|[(3a+2)/(5a+4)]-(3/5)| < ζ\\ \Leftrightarrow&|-2/5(5a+4)| < ζ\\ \Leftrightarrow& a > 1/5[(2/5ζ)-4]=∆ \end{align}$$ hence for every $ζ>0$ there exists $∆>0$ such that which satisfies the definition of convergence hence $f(a)=[(3a+2)/(5a+4)]$ converges to $3/5$.