Vertical asymptote with an epsilon-delta proof?

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In summary, the conversation discusses the aim of proving a vertical asymptote for the function ##\frac{1}{x}## and the difficulties in finding a solution. The conversation mentions using the ordering of real numbers to show that the function is strictly decreasing and has no upper bound, leading to a vertical asymptote at zero. A proof is provided using a delta-epsilon proof, but it is noted that it is not a complete proof and more structure is needed.
  • #1
mcastillo356
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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.

Vertcal asymptote.jpg
Greetings!
 
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  • #2
Doesn't this follow directly from the ordering of the real numbers? From that you have that

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

EDIT: There is a vertical asymptote at zero for essentially the same reasons that [itex]x^{-1} \to 0[/itex] as [itex]x \to \infty[/itex]; it is easy to see that the curve [itex](x,x^{-1})[/itex] is symmetric about the line [itex](x,x)[/itex].
 
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  • #4
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!
 
  • #5
mcastillo356 said:
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.
 
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