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Math Help - Determine where f has a limit

  1. #1
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    Determine where f has a limit

    Define f:R--->R as follows:
    f(x)=x-[x] if [x] is even
    f(x)=x-[x+1] if [x] is odd

    Determine the points where f has a limit and justify

    Not quite sure how to get started...
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  2. #2
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    Quote Originally Posted by kathrynmath View Post
    Determine the points where f has a limit and justify. Not quite sure how to get started...
    Just draw a graph.
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  3. #3
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    So, I just use the graph and see where ther is a limit? Then I can use a formula to justify that these are correct limits?
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    Quote Originally Posted by kathrynmath View Post
    So, I just use the graph and see where ther is a limit? Then I can use a formula to justify that these are correct limits?
    YES And here is a graph.
    Attached Thumbnails Attached Thumbnails Determine where f has a limit-un123.gif  
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  5. #5
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    Ok, so there is a limit everywhere except for the integers. how would I justify using a theorem. Lim(F(x)+G(x))=LimF(x)+LimG(x)?
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  6. #6
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    Quote Originally Posted by kathrynmath View Post
    Ok, so there is a limit everywhere except for the integers. how would I justify using a theorem. Lim(F(x)+G(x))=LimF(x)+LimG(x)?
    Never mind I got the first part justified. then for the 2nd part, i think I find 2 sequences that converge to the same limit, but f(x_n) doesn't converge to the same limit. hence limit DNE.
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  7. #7
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    Quote Originally Posted by kathrynmath View Post
    Ok, so there is a limit everywhere except for the integers. how would I justify using a theorem. Lim(F(x)+G(x))=LimF(x)+LimG(x)?
    Is there a limit everywhere except at the integers. Do I consider each case seperately where [x} is even and the other case?
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  8. #8
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    Limit only at the even numbers?
    if 1<x<2, [x]=1 Then f(x)=x-2. The limit=0
    If 2<x<3 [x]=2 Then f(x)=x-2 The limit =0

    a-1<x<a where a is an odd number. [x]=a-1. Then f(x)=x-(a-1). Limit as x goes to a=1
    a<x<a+1 where a is an odd number. [x]=a. Then f(x)=x-(a-1+1). Limit as x goes to a= 0
    a-1<x<a where a is an even number. [x]=a-1. Then f(x)=x-(a-1+1). Limit=0
    a<x<a+1 where a is an even number. [x]=a. Then f(x)=x-a. Limit = 0

    So the limit exists for even numbers, it seems.
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  9. #9
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    Quote Originally Posted by kathrynmath View Post
    Limit only at the even numbers?
    if 1<x<2, [x]=1 Then f(x)=x-2. The limit=0
    If 2<x<3 [x]=2 Then f(x)=x-2 The limit =0

    a-1<x<a where a is an odd number. [x]=a-1. Then f(x)=x-(a-1). Limit as x goes to a=1
    a<x<a+1 where a is an odd number. [x]=a. Then f(x)=x-(a-1+1). Limit as x goes to a= 0
    a-1<x<a where a is an even number. [x]=a-1. Then f(x)=x-(a-1+1). Limit=0
    a<x<a+1 where a is an even number. [x]=a. Then f(x)=x-a. Limit = 0

    So the limit exists for even numbers, it seems.
    Is this the right idea?
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