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Prove that $\lim_{n\rightarrow \infty} \frac{3n+5}{2n+7} = \frac{3}{2}$ through epsilon delta

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Prove that $\lim_{n\rightarrow \infty} \frac{3n+5}{2n+7} = \frac{3}{2}$Proof:

Let $a_n = \frac{3n+5}{2n+7}$. Then, $\left | a_n-\frac{3}{2} \right |=\frac{11}{2(2n+7)}<\frac{3}{n}$. Given $\epsilon > 0$, choose $n_0 \in \mathbb{N}$ such that $n_0 > \frac{3}{\epsilon}$. Then, for all $n \ge n_0, \left | a_n-\frac{3}{2} \right | < \epsilon $. Therefore, $\lim a_n = \frac{3}{2}$.

I'm not quite sure I understand this proof. Namely:

  1. Where did $\frac{3}{n}$ come from?
  2. I guess we just replaced n with $\epsilon$ in the third sentence.
  3. How did the fourth sentence follow from the third, and why does that show that 3/2 is the limit?

So I tried it myself on another problem, where $a_n = \frac{2n+5}{6n-3}$, and the limit being $\frac{1}{3}$:

Let $a_n = \frac{2n+5}{6n-3}$. Then, $\left | a_n-\frac{1}{3} \right |=\left | \frac{6n-6n+15-3}{3(6n-3)} \right |= \left | \frac{12}{3(6n-3)} \right |= 4\left | \frac{1}{6n-3} \right | < \epsilon$. Dividing by 4, $\left | \frac{1}{6n-3} \right | < \frac{\epsilon}{4}$.

And that's as far as I logically get... I assume the next step is to choose an $n_0 \in \mathbb{N}$ such that $n_0 > \frac{\epsilon}{4}$, and isolating epsilon, $\epsilon < 4n_0$.

but as you can tell, I'm really lost.


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