<s>[INST] Prove $\lim_{a \to \infty} \frac{3a+2}{5a+4}= \frac35$ using definition of limit 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?
 [/INST] 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 $3/5$.
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