Strange limit question

  • Thread starter jdstokes
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  • #1
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Let [itex]f: \mathbb{R} \rightarrow \mathbb{R}[/itex] be given. Let [itex]L[/itex] be a real number. State the condition for saying that as [itex]x[/itex] tends to [itex]a[/itex], the limit of [itex]f(x)[/itex] is not [itex]L[/itex]. The statement ought to begin with "Given there exists [itex]\epsilon > 0[/itex]".

Best guess: [itex]\lim_{x \rightarrow a}f(x) \neq L[/itex] means, given there exists [itex]\epsilon > 0[/itex] there exists [itex]\delta > 0[/itex] such that [itex]|x-a|<\delta \Rightarrow |f(x) - L| > \epsilon[/itex]. I'm really not sure about this, however.
 

Answers and Replies

  • #2
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Nonremovable singularity?
 
  • #3
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Pardon me?
 
  • #4
saltydog
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First, lets go over the rule for the limit:

Given [itex]\epsilon>0[/itex], there exists a [itex]\delta>0[/itex] such that for all x satisfying:

[tex]|x-a|<\delta[/tex]

we have:

[tex]|f-L|<\epsilon[/tex]


Now, what happens if L is not the limit?


Then there should exists an [itex]\epsilon>0[/itex] such that for some [itex]\delta>0[/itex] and all x satisfying:


[tex]|x-a|<\delta[/tex]

we have:

[tex]|f-L|>\epsilon[/tex]

Make sure you understand these two differences. If L is the limit, by choosing [itex]\delta [/itex] small enough I can make the difference between L and f as small as I want.

If L is not the limit, then no matter how small I make [itex]\delta[/itex], I can always find an [itex]\epsilon[/itex] such that the difference between f and L will be larger.
 
  • #5
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So I was correct? Great, thanks for your help saltydog.
 

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