Delta epsilon proof with $x \rightarrow \infty$ I have a rational function that reaches a horizontal asymptote as $x \rightarrow \infty$.
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
 A: 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 $17/(2\epsilon)$.
A: 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 $-\infty$.
A: 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 $x$.
A: 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.
