Proof: limit does not exist w/ delta-epsilon definition

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Vertciel

Junior Member
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May 13, 2007
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Hello everyone

I would like to use the formal definition of a limit to prove that a limit does not exist. Unfortunately, my textbook
(Calculus: One Variable by Salas) does not offer any worked examples involving the following type of limit
so I am not sure what to do.

Thank you for your help.

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\(\displaystyle & \text{Prove that }\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\text{ does not exist}\text{.}\)

I know that I must negate the limit definition, as such:
\(\displaystyle \\ &\forall L : \exists \epsilon > 0 : \forall \delta > 0 : \exists x : \left(0 < \left|x-c\right| < \delta\right) \,\, \Rightarrow \,\, \left( \left| f(x) - L \right| > \epsilon\right)\)
Also, I believe that I must involve the formal definition of the limit on \(\displaystyle (0,1]\) since this is where \(\displaystyle f(x) =\frac{1}{x}\)
increases without bound.
However, how would I go about doing this?
 
Take some L and let that be the limit. As x approaches 0, then the value of 1/x should be within epsilon of L.

But, as x--->0, 1/x--->infinity. Therefore, epsilon gets bigger and bigger, not smaller and smaller as it should.

So, regardless of what L you pick, when x is near a, some f(x) values are NOT near L? Then the limit does not exist. As in thia case.

Go here and enter in 1/x as the new function, and set limit at a equal to 0. Enter in whatever for the test limit. This applet does a fine job of allowing one to see what is going on.

http://www.slu.edu/classes/maymk/Applet ... Delta.html
 
Thank you for your response, galactus.

I understand the intuition behind your post and want to express it as a formal proof. I am still not sure how to do this, but I tried to understand my instructor's proof below, which I unfortunately don't:

Instructor's Proof:

Let L be in the set of real numbers.
Define \(\displaystyle \epsilon = 1\). Let \(\displaystyle \delta > 0\).
If \(\displaystyle \abs{x - 0} < \delta\), then \(\displaystyle |\frac{1}{x} - L| \geq \epsilon \Rightarrow |\frac{1}{x} - L| \geq 1\)

For \(\displaystyle L \geq 0\): Choose \(\displaystyle 0 < x < \min({\delta, \frac{1}{L+1}})\) -- (2)

For \(\displaystyle L < 0\): Choose \(\displaystyle 0 < x < \min({\delta, \frac{1}{\abs{L}+1}})\) --(3)

1. Could you please explain from where (2) and (3) came?

2. How does my instructor's proof state rigorously that as \(\displaystyle x \rightarrow 0\), no value of delta will give an epsilon which will lie within a certain interval including L.
 
\(\displaystyle Another, \ way; \ proof \ by \ contradiction, \ which \ should \ be \ easier \ to \ understand.\)

\(\displaystyle Statement: \ \lim_{x\to0}\frac{1}{x} \ does \ not \ exist. \ Suppose \ not.\)

\(\displaystyle Suppose \ \lim_{x\to0}\frac{1}{x} \ = \ L, \ L \ a \ real \ number \ and \ let \ \epsilon \ = \ 1 \ and \ \delta \ > \ 0.\)

\(\displaystyle Then, \ \bigg|\frac{1}{x}-L\bigg| \ < \ 1 \ whenever \ 0 \ < \ |x-0| \ < \ \delta; \ in \ particular, \ let \ x \ = \ \frac{\delta}{\delta+1} \ and \ x \ = \ \frac{-\delta}{\delta+1}\)

\(\displaystyle Note: \ \bigg|\frac{\delta}{\delta+1}-0\bigg| \ < \ \delta \ and \ \bigg|\frac{-\delta}{\delta+1}-0\bigg| \ < \ \delta\)

\(\displaystyle Hence, \ \bigg|\frac{1}{x}-L\bigg| \ < \ 1 \ \implies \ \bigg|\frac{\delta+1}{\delta}-L\bigg| \ < \ 1 \ and \ \bigg|\frac{-\delta-1}{\delta}-L\bigg| \ < \ 1,\)

\(\displaystyle or \ equivalently, \ \frac{1}{\delta} \ < \ L \ < \ \frac{1}{\delta}+2 \ and \ -2-\frac{1}{\delta} \ < \ L \ < \ -\frac{1}{\delta}\)

\(\displaystyle which \ is \ impossible \ (limits \ are \ unique). \ Therefore \ the \ supposition \ is \ false \ and \ the \ original\)

\(\displaystyle statement \ is \ true. \ QED\)
 
BigGlenntheHeavy said:
\(\displaystyle Another, \ way; \ proof \ by \ contradiction, \ which \ should \ be \ easier \ to \ understand.\)

\(\displaystyle Statement: \ \lim_{x\to0}\frac{1}{x} \ does \ not \ exist. \ Suppose \ not.\)

\(\displaystyle Suppose \ \lim_{x\to0}\frac{1}{x} \ = \ L, \ L \ a \ real \ number \ and \ let \ \epsilon \ = \ 1 \ and \ \delta \ > \ 0.\)

\(\displaystyle Then, \ \bigg|\frac{1}{x}-L\bigg| \ < \ 1 \ whenever \ 0 \ < \ |x-0| \ < \ \delta; \ in \ particular, \ x \ = \ \frac{\delta}{\delta+1} \ and \ x \ = \ \frac{-\delta}{\delta+1}\)

\(\displaystyle Note: \ \bigg|\frac{\delta}{\delta+1}-0\bigg| \ < \ \delta \ and \ \bigg|\frac{-\delta}{\delta+1}-0\bigg| \ < \ \delta\)

\(\displaystyle Hence, \ \bigg|\frac{1}{x}-L\bigg| \ < \ 1 \ \implies \ \bigg|\frac{\delta+1}{\delta}-L\bigg| \ < \ 1 \ and \ \bigg|\frac{-\delta-1}{\delta}-L\bigg| \ < \ 1,\)

\(\displaystyle or \ equivalently, \ \frac{1}{\delta} \ < \ L \ < \ \frac{1}{\delta}+2 \ and \ -2-\frac{1}{\delta} \ < \ L \ < \ -\frac{1}{\delta}\)

\(\displaystyle which \ is \ impossible \ (limits \ are \ unique). \ Therefore \ the \ supposition \ is \ false \ and \ the \ original\)

\(\displaystyle statement \ is \ true. \ QED\)
Click to expand...

Thank you for your response, BigGlenntheHeavy.

Would you mind explaining how you got \(\displaystyle \ x \ = \ \frac{\delta}{\delta+1} \ and \ x \ = \ \frac{-\delta}{\delta+1}\)?
 
BigGlenntheHeavy said:
\(\displaystyle Then, \ \bigg|\frac{1}{x}-L\bigg| \ < \ 1 \ whenever \ 0 \ < \ |x-0| \ < \ \delta; \ in \ particular, \ x \ = \ \frac{\delta}{\delta+1} \ and \ x \ = \ \frac{-\delta}{\delta+1}\)

[/tex]
Click to expand...

Thank you for your response, BigGlenntheHeavy.

Would you mind explaining how you got \(\displaystyle \ x \ = \ \frac{\delta}{\delta+1} \ and \ x \ = \ \frac{-\delta}{\delta+1}\)?[/quote]
 
\(\displaystyle If \ any \ a, \ b, \ c, \ \delta, \ is\ > \ 0, \ then \ \frac{a}{a+1}, \ \frac{b}{b+1}, \ etc., \ is \ < \ than \ a, \ b, \ etc.\)

\(\displaystyle Note: \ let \ \delta \ = \ 5, \ then \ \frac{5}{6} \ < \ 5, \ let \ \delta \ = \ \frac{1}{100}, \ then \ \frac{1}{101} \ < \ \frac{1}{100}.\)

Addendum: When doing delta-epsilons, it behooves one to put on his thinking cap.

You can let x = anything as long as it is less than delta, I chose the above two to show that no limit existed.
 
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