Prove that the sequence $(3n^2+4)/(2n^2+5) $ converges to $3/2$ 
Prove directly from the definition that the sequence $\left( \dfrac{3n^2+4}{2n^2+5} \right)$ converges to $\dfrac{3}{2}$.

I know that the definition of a limit of a sequence is $|a_n - L| < \varepsilon$ .
However I do not know how to prove this using this definition. 
Any help is kindly appreciated. 
 A: Let $\epsilon>0$,
$$\left|u_n-\frac{3}{2}\right|=\left|\frac{3n^2+4}{2n^2+5}-\frac{3}{2}\right|=\frac{7}{4n^2+10}<\epsilon\iff 4n^2>\frac{7}{\epsilon}-10$$
so let $N=\frac{1}{2}\sqrt{\left|\frac{7}{\epsilon}-10\right|}$ then forall $n>N$ we have $\left|u_n-\frac{3}{2}\right|<\epsilon$ and we conclude.
A: $$\left|\frac{3n^2+4}{2n^2+5}-\frac{3}{2}\right| = \left|\frac{6n^2+8-6n^2-15}{4n^2+10}\right| = \frac{7}{4n^2+10}$$
How would we continue?
A: Hints: for an arbitrary $\,\epsilon>0\;$ :
$$\left|\frac{3n^2+4}{2n^2+5}-\frac32\right|=\left|\frac{-7}{2(2n^2+5)}\right|<\epsilon\iff2n^2+5>\frac7{2\epsilon}\iff$$
$$(**)\;\;2n^2>\frac7{2\epsilon}-5\ldots$$
Be sure you can prove you can choose some $\,M\in\Bbb N\;$ s.t. for all $\,n>M\;$ the inequality (**) is true.
A: Suppose that it does not converge to $3/2$. Then you have that there exists $\varepsilon_0>0$ such that for every $n\in\mathbb N$ we have $|\displaystyle \frac {3n^2+4}{2n^2+5} - \frac {3}{2}|\geq \varepsilon_0 $ which is equivalent to $\displaystyle \frac {7}{4n^2+10}\geq \varepsilon_0$ for every $n\in\mathbb N$ which is not true since we can make the fraction as small as we like so the sequence is convergent and the limit is $3/2$.
A: The following computation proves the claim
$$\left|\frac{3}{2}-\frac{3n^2 + 4}{2n^2+5}\right| = \left|\frac{(6n^2+15)-(6n^2+8)}{4n^2+10}\right| = \left|\frac{7}{4n^2+10}\right|\stackrel{n\rightarrow\infty}{\longrightarrow}0$$
The same computation shows that in general if you have two polynomials $P$ and $Q$ of the same degree $m$, then $\frac{P(n)}{Q(n)}$ converges to $\frac{p_m}{q_m}$, where $p_m$ is the coefficient of $x^m$ in $P$ and similarly for $Q$.
