<s>[INST] Find the following limit and prove its value by definition. $\lim _{n\to \infty} (n^2 + 3n + 7)/(2n^2 −5n − 4)$. Find the following limit and prove its value by definition. $\lim_{n\to \infty} \frac{n^2 + 3n + 7}{2n^2 −5n − 4}$. I am supposed to use$$ \left|\frac{n^2 + 3n + 7}{2n^2 −5n − 4} - \frac{1}{2}\right| < ε $$then combine the fraction of the limit and the result and eventually get rid of the absolute value bars with some algebra and end up with one $n <$ some $ε$ but since neither the top nor bottom of the limit factor I do not know how I can manipulate the limit so my final inequality can be a single $n$ and some $ε$.
 [/INST] Since
$$\frac{n^2 + 3n +7}{2n^2-5n+4} - \frac{1}{2} = \frac{2n^2 + 6n +14 - 2n^2+5n-4}{4n^2-10n-8}  = \frac{11n-10}{4n^2-10n-8} $$
Then you could simply bound the right hand side,
$$|\frac{11n-10}{4n^2-10n-8} | \leq |\frac{11n}{4n^2-10n-8}| \leq |\frac{11n}{4n^2-11n}| = |\frac{11}{4n-11}| \leq |\frac{11}{4n-n}|  \leq \frac{11}{3n}$$
Where we used that $n$ has to bigger than $8$ and $11$, thus the above inequality holds for $n \geq 11$. We want, given $\varepsilon >0$, find an $n \in \mathbb N$ such
$$|\frac{11n-10}{4n^2-10n-8} |<\varepsilon,$$
but we can bound the upper term, in the following way:
$$|\frac{11n-10}{4n^2-10n-8} |< \frac{11}{3n}<\varepsilon$$
for $n \geq 11$. So we only have to find the $n$ that makes
$$\frac{11}{3n}<\varepsilon$$
true. But the previous inequality is equivalent to
$$ \frac{11}{3 \varepsilon} < n. $$ This means the inequality is true of $n$'s bigger than $\frac{11}{3 \varepsilon}$. We can take, for example, $n = \lceil \frac{11}{3 \varepsilon} \rceil + 1$. But don't forget a condition is also that $n \geq 11$! So finally,
$$n_0 = \max\left\{11, \left \lceil \frac{11}{3 \varepsilon} \right \rceil + 1 \right\} $$ is an $n$ that makes $|\frac{11n-10}{4n^2-10n-8} |<\varepsilon$ true.
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