<s>[INST] Prove $\lim_{n\to\infty}{\frac{3n+1}{2n+5}=\frac{3}{2}}$ Problem Prove $$\lim_{n\to\infty}{\frac{3n+1}{2n+5}=\frac{3}{2}}$$ using a epsilon proof
Scratchwork
We see that $\mid{\frac{3n+1}{2n+5}-\frac{3}{2}}\mid=\mid{\frac{6n+2}{4n+10}-\frac{6n+15}{4n+10}}\mid= \mid\frac{-13}{4n+10}\mid=\frac{13}{4n+10}<\frac{13}{4n}<\frac{13}{n}<\epsilon$.
(I like to go all the way to $\frac{13}{n}$ which im not sure i can do). If we let $\epsilon >0$ then $\frac{1}{\epsilon}>0\implies \frac{13}{\epsilon}$ By the Archimdean property we can choose a $k\in\mathbb{N}$ such that $\frac{13}{\epsilon}<k\implies \frac{13}{k}<\epsilon$. 
Proof:
Let let $\epsilon >0$ then $\frac{1}{\epsilon}>0\implies \frac{13}{\epsilon}>0$ Then by the Archimedean property there exists a $k\in\mathbb{N}$ such that $k>\frac{13}{\epsilon}\implies \frac{13}{k}<\epsilon$. If $n \geq k$ then $\frac{1}{n} \leq \frac{1}{k}\implies \frac{13}{n}\leq \frac{13}{k} <\epsilon$. Thus$\mid{\frac{3n+1}{2n+5}-\frac{3}{2}}\mid<\frac{13}{n}<\epsilon$. Thus $\lim{\frac{3n+1}{2n+5}=\frac{3}{2}}$. Would this suffice? 
 [/INST] Yes, absolutely :)
Would post this as a comment, but then this question remains unanswered, so... 
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