Spivak Chapter 5 Problem 26) a

  • #1
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Homework Statement


Give an example to show that the given "definition" of limx→aƒ(x) = L is incorrect.

Definition: For each 0<δ there is an 0<ε such that if 0< l x-a I < δ , then I ƒ(x) - L I < ε .

Homework Equations




The Attempt at a Solution


I considered the piece-wise function: ƒ(x) = (0 if x<0) = ( λ if x>0). I then chose an ε such that ε > I λ/2 I , and chose L = λ/2.
It is obviously true that for each positive δ , if 0 < I x-0 I < δ , then I ƒ(x) - λ/2 I < ε . But, by our definition, this means that limx→0ƒ(x) = λ/2 , which is blatantly false.

Is this sufficient? I think that spivak is expecting that I use some sort of "common sense" in my argument such as in my final statement.
 

Answers and Replies

  • #2
Dick
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Homework Statement


Give an example to show that the given "definition" of limx→aƒ(x) = L is incorrect.

Definition: For each 0<δ there is an 0<ε such that if 0< l x-a I < δ , then I ƒ(x) - L I < ε .

Homework Equations




The Attempt at a Solution


I considered the piece-wise function: ƒ(x) = (0 if x<0) = ( λ if x>0). I then chose an ε such that ε > I λ/2 I , and chose L = λ/2.
It is obviously true that for each positive δ , if 0 < I x-0 I < δ , then I ƒ(x) - λ/2 I < ε . But, by our definition, this means that limx→0ƒ(x) = λ/2 , which is blatantly false.

Is this sufficient? I think that spivak is expecting that I use some sort of "common sense" in my argument such as in my final statement.
It seems sufficient to me. You've taken a function that looks discontinuous and shown it fits the 'definition'.
 
  • #3
PeroK
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Homework Statement


Give an example to show that the given "definition" of limx→aƒ(x) = L is incorrect.

Definition: For each 0<δ there is an 0<ε such that if 0< l x-a I < δ , then I ƒ(x) - L I < ε .

Homework Equations




The Attempt at a Solution


I considered the piece-wise function: ƒ(x) = (0 if x<0) = ( λ if x>0). I then chose an ε such that ε > I λ/2 I , and chose L = λ/2.
It is obviously true that for each positive δ , if 0 < I x-0 I < δ , then I ƒ(x) - λ/2 I < ε . But, by our definition, this means that limx→0ƒ(x) = λ/2 , which is blatantly false.

Is this sufficient? I think that spivak is expecting that I use some sort of "common sense" in my argument such as in my final statement.
Your example is valid, but you may be missing the point somewhat.

First, when finding an example, you can (and perhaps it's better to) use some definite numbers. E.g. why not just have ##\lambda = 1##?

Also, very strictly speaking, you didn't actually find an ##\epsilon##. This does not mean what you did was wrong. But, why not give a specific ##\epsilon##? With ##\lambda = 1##, you could have taken ##\epsilon = 1##. Or, in your general case ##\epsilon = |\lambda|##.

The reason I mention this is that a reluctance (or inability) to choose a specific ##\epsilon## or ##\delta## can lead to difficulties in finding counterexamples or proving a limit does not exist.

Also, to show that the definition is not just invalid for unusual functions, I suggest finding another example for the function:

##\forall x \ f(x) = 0##
 

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