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Limits and choosing an epsilon properly?

  • Thread starter matrix_204
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  • #1
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I really need help on solving this question:

Let d and K be given real numbers. Suppose that lim f(x) > K.
x->c
Show that there is a number h>0 such that f(x) > K for all x in the punctured open interval of width 2h centred at d.

The only hint that i was given was that if there are two real numbers as close as you like, then they are basically the same real.
How can i show this using this idea?
 

Answers and Replies

  • #2
arildno
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Is c=d?????????????
 
  • #3
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yes that is the idea, but how am i suppose to show that it's equal
 
  • #4
shmoe
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Suppose [tex]\lim_{x\rightarrow c}f(x)=L>K[/tex]

Write down the definition of the limit in this case. There's a [tex]|f(x)-L|<\epsilon[/tex] part. This controls how close f is to L. By chosing epsilon properly, you can force f to be some distance away from any number not equal to L (on some punctured disc centered at c of course).

For example if you know [tex]|f(x)-L|<1/2[/tex] on some interval, then can [tex]f(x)=L+1/2[/tex] on this interval? Can it equal anything larger? What's the lowest it could be?
 
  • #5
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What does it mean when asking to chosing an epsilon properly?



Also, can someone clarify, how the graph of two horizontal lines work? i.e, y=L , y=f.
 
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  • #6
shmoe
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matrix_204 said:
What does it mean when asking to chosing an epsilon properly?

Also, can someone clarify, how the graph of two horizontal lines work? i.e, y=L , y=f.
You can think of epsilon as bounds for your function on the corresponding interval [tex]0<|x-c|<\delta[/tex].On this interval, your function will only take on values above [tex]L-\epsilon[/tex] and below [tex]L+\epsilon[/tex]. How do you pick epsilon to leave K out of this range? If epsilon is too large, you won't be able to rule out the possibility that f(x)=K.


Your horizontal lines..y=L would just be a horizontal line at height L, y=f won't necessarily be a horizontal line, f is a function. I don't think I understand your question.
 
  • #7
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So in order to show that f(x)> K for all x.. what are the main steps required in proving this, and also is there any techniques that can be used in solving delta-epsilon type of problems? I m really confused in these types of problems, i dont seem to understand the concept of delta and epsilon, like i kno that they are really small and can be regarded as equal, or not equal but very close, etc.
 
  • #8
shmoe
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1) pick an [tex]\epsilon >0[/tex] so that if [tex]|f(x)-L|<\epsilon[/tex] you know [tex]|f(x)-K|>0[/tex]. Your particular epsilon will depend on how far L is from K.
2) Appeal to the definition of the limit to produce a [tex]\delta >0[/tex] so that if [tex]0<|x-c|<\delta[/tex] then [tex]|f(x)-L|<\epsilon[/tex]. You won't know what this delta is, but the fact that the limit is L guarantees it's existance.
3) Combine the above.

epsilon-delta type proofs are important if you want an understanding of calculus. Everything you do later on depends on limits, and without understanding epsilon-delta you won't rigorously understand what a limit is. You really need to work some examples and try to understand the definition of the limit as much as possible. Practice is important.
 

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