$\epsilon-N$ for $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$.

Question

I want to prove using $\epsilon-N$ that $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$.

Firstly we will find a sufficently large $n\in \Bbb N$:

\begin{align*} \epsilon\quad\gt&\quad\left|\frac{3n+1}{2n+5} - \frac32 \right|\\ =&\quad \left|\frac{3n+1 -3n-7.5}{2n+5}\right|\\ =&\quad \left|\frac{-13}{4n+10}\right|\\ \implies&\quad \epsilon \gt \frac{13}{4n+10}\\ \implies&\quad \frac1\epsilon \lt \frac{4n+10}{13}\\ \implies&\quad \frac14\left(\frac{13}\epsilon-10\right) \lt n\\ \end{align*}

So $\forall \epsilon\gt 0, \exists N(\epsilon)>\frac14\left(\frac{13}\epsilon-10\right)$ such that $\forall n\in N, n>N$, $a_n=|\frac{3n+1}{2n+5}-\frac32|\lt \epsilon$

Proof: $\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right| \color{red}{\lt} \left|\frac{3N+1}{2N+5}-\frac{3}{2}\right|\lt\epsilon$

Assuming this is even right. Everything feels good except the $\color{red}{\lt}$ in red, and then when I check, since we have

\begin{align*} &\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right| \color{red}{\lt}\left|\frac{3N+1}{2N+5}-\frac{3}{2}\right|,\quad n\gt N\\ &\implies \left|\frac{3n+1}{2n+5} - \frac32\right| - \left|\frac{3(n+1)+1}{2(n+1)+5} - \frac{3}{2}\right|\gt 0 \end{align*}

But this fails for all $n$? What did I do wrong?

Also, since $N\in\Bbb N$, we have to choose our $N$ so it is a natural number simultaneous to being greater than $\frac14\left(\frac{13}\epsilon-10\right)$, how do I ensure that subbing in $\frac14\left(\frac{13}\epsilon-10\right)$ is greater than subbing in the natural number that is the ceiling of this?

Answer 1

What is the problem?

$\large\mid{3n+1\over2n+5}-{3\over2}\mid={13\over4n+10}$

$\large\mid{3N+1\over2N+5}-{3\over2}\mid={13\over4N+10}$

$\large\mid{3(n+1)+1\over2(n+1)+5}-{3\over2}\mid={13\over4(n+1)+10}$

So for $n>N$ the second one is obviously larger than the first one and the first one is obviously larger than the third one.

Answer 2

$\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$.

Let $\epsilon >0$ be arbitrary.

Now by the archemedian property there exists $N \in \mathbb N$ such that $N>\frac{13}{\epsilon}$.

Observe that for each $n>N $

$|\frac{3n+1}{2n+5}-\frac{3}{2}|=\frac{13}{4n+10}< \frac{13}{n}<\frac{13}{N}<\epsilon$.