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?
 A: Your proof seems fine to me. It seems clear from what you wrote that you know what you are doing. And I like that at the beginning of your proof you clearly stated what is your goal. 
Still they are a few minor points. (You may call this nitpicking if you wish.)


*

*The symbol $\Rightarrow$ is used to denote impication. I do not like the place where you wrote
$$\Rightarrow \left|\frac{3a+2}{5a+4} - \frac{3}{5}\right|\ = \dots$$
because what you wrote here is in fact not a consequence of what you wrote previously. This is the place where you are trying to simplify the expression you need to estimate. (So I would simply omit $\Rightarrow$ at that place.)

*The inequality $$\left|\frac{-2}{5(5a+4)}\right|\le\frac{1}{a}$$ is more-or-less clear, but you might add an explanation why this inequality holds. (I'd say this depends on who is grading this, if it is an assignment. Different people might expect different level of detail. The most important thing is that you understand why the inequality used in your proof holds.)


Since it is possible that you will further edit your post, I will add a link to the revision for which my comments were intended.
A: 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 $3/5$.
