<s>[INST] Show $\lim\limits_{n\to\infty} \frac{2n^2-3}{3n^ 2+2n-1}=\frac23$ Using Formal Definition of Limit I want to show that $a_n=\frac{2n^2-3}{3n^ 2+2n-1}$ is convergent. So I did the following:
\begin{align*}
\left|a_n-\frac23\right|&=\left|\frac{2n^2-3}{3n^ 2+2n-1}-\frac23\right|\\
&=\left|\frac{-4n-7}{3(3n^2+2n-1)}\right| \\
&<\left|\frac{4n}{3n^2}\right|\tag{$\ast$}\\
&<\left|\frac4n\right|\\
&<\frac4N\\\
\end{align*}
But I am not one hundred percent sure about ($\ast$) because $|-4n-7|=|4n+7|\not<4n$. 
Can somebody please explain my error in reasoning?
 [/INST] Suppose we consider $n \geq 1$. Then,
\begin{align*}
\left|a_n-\frac23\right|&=\left|\frac{2n^2-3}{3n^ 2+2n-1}-\frac23\right|\\
&=\left|\frac{-4n-7}{3(3n^2+2n-1)}\right| \\
&=\frac{4n+7}{3(3n^2+2n-1)}.
\end{align*}
You may now proceed by saying $4n + 7 \leq 5n,\,n\geq 7$, and $3n^2+2n-1\geq 3n^2,\,n\geq 1$, and therefore,
\begin{align*}
\left|a_n-\frac23\right|\leq \frac{5}{9n},\,n\geq 7.
\end{align*}
Regarding your reasoning, you need to actually show that (you have not done it)
\begin{align*}
\frac{4n+7}{3(3n^2+2n-1)} \leq \frac{4}{3n}
\end{align*}
This holds if and only if (cross multiplying after canceling the common $3$s in the denominators)
$$4n^2+7n \leq 12n^2 + 8n-4,$$
which indeed holds for sufficiently large $n$.
Even though both methods work, the simplest (and the less error-prone) way is to treat the numerator and the denominator separately.
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