Delta epsilon proof with x approaches infinity

I have a rational function that reaches a horizontal asymptote as x approaches infinity. How would you do a delta-epsilon proof with $x\to\infty$.

Here is the limit statement:

$$\lim_{x\to\infty}\frac{3x+7}{2x-1} = \frac{3}{2}.$$

Hope some one can help.

Paulo

Hint: when you use $\epsilon$-$\delta$ definition of the convergence $\lim_{x\to\infty }f(x) = a$, it looks like $$ \forall \epsilon>0\;\exists\delta\in\mathbb R: \;x>\delta\implies |f(x) - a|<\epsilon. $$ In your case $f(x) = \frac{3x+7}{2x-1}$ and $a = \frac32$, so that for any $\epsilon>0$ you have to found $\delta$ big enough s.t. $$ \left|\frac{3x+7}{2x-1} - \frac32\right|<\epsilon \tag{1} $$ for all $x>\delta$. Namely, I would suggest you solve the inequality $(1)$ in terms of $x$.

Given any $\epsilon>0$, we want to find $M$ such that

$$\left|\frac{3x+7}{2x-1}-\frac{3}{2}\right|<\epsilon$$

whenever $x>M$. Combining fractions, we get:

$$\left|\frac{17}{4x-2}\right|<\left|\frac{17}{2x}\right|$$

Now, if we want to be less than $\epsilon$, so if we assume it is, we get $x>17/(2\epsilon)$. So let $M$ be anything greater than $17/(2\epsilon)$.

Normally, proofs as something approaches infinity are not framed as $\epsilon$-$\delta$ proofs at all. $\delta$ ends up getting replaced by some other letter. However... the extended real line is homeomorphic to $[0,1]$, so you could impose the corresponding metric to it, and then you'd have a meaningful notion of how far a given number is from $\infty$ or from $-\infty$.

It will be much easier to do this proof after going through polynomial division. In other words, we write that $$ \frac{3x+7}{2x-1}=\frac{(3/2)(2x-1)+17/2}{2x-1}=\frac32+\frac{17}{2(2x-1)} $$ After that, the proof just amounts to showing that $f(x)=\frac{17}{2(2x-1)}$ approaches zero. That is, we want to show that $\forall \epsilon>0\;\exists\delta>0: \;x>\delta\implies |f(x)|<\epsilon$. Where we would normally have a $|x-c|<\delta$, we now just have $x>\delta$ since our goal is to show that the larger $x$ gets (i.e. the closer to infinity), the smaller $f$ (our difference) gets.

In fact, selecting $\delta=\max\{\frac14(2+17/\epsilon),1\}$ and noting that $f$ is decreasing over $x>1/2$, we have the inequality: $$ |f(x)|=\left|\frac{17}{2(2x-1)}\right|=\frac{17}{2(2x-1)}\\ <\frac{17}{2(2(\frac14(2+17/\epsilon))-1)}=\cdots=\epsilon\\ $$

That completes our proof.