precise definition of limit at infinity I am supposed to prove this limit using the Precise Definition of limit.
$$\lim_{x\to \infty}\frac{3x+2}{2x+3}=\frac32$$
Given $\epsilon>0$, there is a number $N$ such that $x>N$ implies that \begin{eqnarray}
\left|\frac{3x+2}{2x+3}-{3/2}\right| &< & \epsilon, \\
\implies \left|\frac{2(3x+2)-3(2x+3)}{2(2x+3)}\right| &<& \epsilon, \\
\implies \left|\frac{-5}{2(2x+3)}\right| &<& \epsilon.
\end{eqnarray}
How do I proceed from here, what should my $N$ be?
 A: For such problems you often need to simplify your inequalities as much as possible. Our goal is to constrain the values of $x$ in a certain manner (namely keeping $x$ greater than an appropriate positive $N$) such that the inequality $$\left|\frac{3x+2}{2x+3}-\frac{3}{2}\right|<\epsilon \tag{1}$$ is satisfied. The last inequality is equivalent to $$\left|\frac{5}{2(2x+3)}\right|<\epsilon $$ If $x>0$ (this is our first constraint on $x$) then the above is equivalent to $$\frac{5}{4x+6}<\epsilon\tag{2}$$ Next note the obvious inequality $$\frac{5}{4x+6}<\frac{5}{x}$$ and hence if we ensure that $5/x<\epsilon$ our desired inequality $(2)$ will automatically be satisfied (this step is the part which one must understand clearly and it is here that we make a great and obvious simplification of our inequalities involved). And therefore our goal is achieved by having $x>5/\epsilon$ (this is our second constraint on $x$). Thus we can see that it is sufficient to choose $N=5/\epsilon$ and then $x>N$ will ensure that the desired inequality $(1)$ will be satisfied.
Problems based on the $\epsilon, \delta$ (or $\epsilon, N$) definition of limit are not supposed to be an exercise in solving inequalities to get something like $x>N$ from the the starting inequality $|f(x) - L|<\epsilon$ via algebraic manipulation. Thinking these problems as algebraic manipulation of inequalities misses the whole point of the definition of limit.
A: Take $$N= \frac{5}{4\epsilon}-\frac32$$
And then proceed by
$$x>N$$
Put value of $N$ and then solve.
A: As the LHS of the last inequality is a decreasing function, you can take any value of $N$ that satisfies it.
$$\frac{5}{2(2N+3)}<\epsilon,$$
then
$$2N+3>\frac{5}{2\epsilon},$$
$$N>\frac12\left(\frac{5}{2\epsilon}-3\right).$$

For such problems, gross estimates are sufficient.
$$\frac{5}{2(2N+3)}<\frac{5}{4N}<\frac2N<\epsilon,$$ and you can take $\dfrac2\epsilon$.
A: One more attempt:
Let $(x_n)_{n \in \mathbb  N} \rightarrow \infty$ :
For every $K \in$  $\mathbb R$ there exists a   $n_0$ such that for $n \ge n_0$   $x_n \ge K$.
$ \epsilon \gt 0$ given. 
Choose $K > \frac{2}{\epsilon}$, and $n \ge n_0$  then: 
$|\frac{(3x_n + 2)}{(2x_n + 3)} - \frac{3}{2}|$ =
$|\frac{ -5}{2(2x_n +3)}| \lt \frac{5}{4x_n} \lt \frac{2}{x_n} \le \frac{2}{K} \lt \epsilon$.
