Vertical asymptote with an epsilon-delta proof?

[mcastillo356]

TL;DR Summary

I've got a very recurred rational function for which I would like to find out the proof of the vertical asymptote.

Hi, PF

The aim is to prove how the approach from the left and right sides of the $x$x axis eventually renders a vertical asymptote for the function $\frac{1}{x}$. I've been searching in the textbook "Calculus", 7th edition, by Robert A. Adams and Christopher Essex, but I haven't found nothing but uncertain clues. Or naive proofs ( i.e., YouTube, mentioning the fact that the more we become near to zero at the abscissa, the less turns the ordinate). Any suggestion would be fine. Apologizes for not attempt provided.


Greetings!

 

[pasmith]

Doesn't this follow directly from the ordering of the real numbers? From that you have that

(1) $x^{-1}$ is strictly decreasing for $x > 0$ ($0 < x < y < 1 \Leftrightarrow 1 < 1/y < 1/x$).
(2) $\{x^{-1} : x > 0 \}$ has no upper bound (consider the sequence $x_n = n^{-1}$ for integer $n > 0$.)
(3) Multiplication by -1 reverses order.

EDIT: There is a vertical asymptote at zero for essentially the same reasons that $x^{-1} \to 0$ as $x \to \infty$; it is easy to see that the curve $(x,x^{-1})$ is symmetric about the line $(x,x)$.

 

[mcastillo356]

Doesn't this follow directly from the ordering of the real numbers?

Thanks a lot! Actually I had seen a proof I thought it was naive; now I realize it is a good demo:

https://www.khanacademy.org/math/ap...-new/ab-1-14/v/infinite-limits-and-asymptotes

Greetings

 

[mcastillo356]

Hi, PF, I've been working with help a delta-epsilon proof of a left-sided limit when $x\rightarrow{0^-}$ for the function $\displaystyle\frac{1}{x}$:

$\forall{\,\epsilon>0}\,\exists{\,\delta>0}$, such that if $-\delta<x<0$, then $\displaystyle\frac{1}{x}<-\epsilon$

If we choose $\delta=\displaystyle\frac{1}{\epsilon}>0$, and substitute in $-\delta<x<0$, we obtain $\displaystyle\frac{1}{x}<-\epsilon<0$. Therefore it is proved that $\lim_{x\rightarrow{0^-}}{\displaystyle\frac{1}{x}}=-\infty$

Is it right? Is $\delta=\displaystyle\frac{1}{\epsilon}$ the solution?

Greetings!

 

[PeroK]

Hi, PF, I've been working with help a delta-epsilon proof of a left-sided limit when $x\rightarrow{0^-}$ for the function $\displaystyle\frac{1}{x}$:

$\forall{\,\epsilon>0}\,\exists{\,\delta>0}$, such that if $-\delta<x<0$, then $\displaystyle\frac{1}{x}<-\epsilon$

If we choose $\delta=\displaystyle\frac{1}{\epsilon}>0$, and substitute in $-\delta<x<0$, we obtain $\displaystyle\frac{1}{x}<-\epsilon<0$. Therefore it is proved that $\lim_{x\rightarrow{0^-}}{\displaystyle\frac{1}{x}}=-\infty$

Is it right? Is $\delta=\displaystyle\frac{1}{\epsilon}$ the solution?

Greetings!

It's the outline of a proof, but not a proof. Also, you begin by stating what it is you want to prove. It's not clear whether this is an assumption (which would be wrong) or a statement of what you are trying to prove.

You must be able to structure an elementary proof correctly before you can tackle more complicated proofs.