<s>[INST] Proving that $\frac{6n^2+3n}{2n^2-5}$ converges to 3. Prove the following:
$$\lim_{n\to \infty}\frac{6n^2+3n}{2n^2-5}=3$$
Here's my solution:
Consider the function $\frac{6n^2+3n}{2n^2-5}$.
Then $$|\frac{6n^2+3n}{2n^2-5}-3|= |\frac{6n^2+3n}{2n^2-5}-\frac{6n^2+15}{2n^2-5}|=|\frac{3n-15}{2n^2-5}|$$
The upper bound for the numerator is $3n-15 \le 3n$ for all $n$ and the lower bound for the denominator is $2n^2-5 \le 4n^2$ if $n \le 2$. 
It follows that 
$|\frac{3n-15}{2n^2-5}| \le \frac{3n}{4n^2}$. Meaning $|\frac{3n-15}{2n^2-5}| \le \frac{3}{4n}$.
By the Archimedean Property, $\frac{3}{4n} \le \frac{3}{4N} \lt \epsilon$
Therefore $(\frac{6n^2+3n}{2n^2-5})\rightarrow 3$.
$\blacksquare$
I feel like I'm missing a step after showing that $\frac{3}{4n} \le \frac{3}{4N} \lt \epsilon$. 
Can $\frac{3}{4n} \le \frac{3}{4N} \lt \epsilon$ be simplified more so that it resembles $\frac{1}{n} \le \frac{1}{N} \lt \epsilon$ or is it fine how it is? 
Please advise. 
 [/INST] To get an upper bound, you want to replace the numerator by something big and the denominator by something small. So instead of $2n^2-5\lt4n^2$, you should use $2n^2-5\gt2n^2$ (or even just $n^2$). Aside from that, everything is basically OK.
</s>