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Math Help - Limits

  1. #1
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    Limits

    If f(x)=x/2 when x belongs to rational numbers
    =0 when x belongs to irrational numbers
    Can limit x reaches 0 exist?

    And If g(x)=0 when x belongs to rational numbers
    =x+1 when x belongs to irrational numbers
    Can limit x reaches 0 exist?

    Can someone please explain about these?
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Limits

    I'll help with the first - similar logic applies to the second. The definition of the limit requires that as x approaches the 0 the value of f(x) gets "closer and closer" to the limit 'L,' meaning that while it doesn't have to actually equal L it gets closer than any arbitrary error (epsilon) that you can name. For this problem you first need to know that for any value of x there are an infinite number of rational and irrational numbers between 0 and x, so the value of f(x) alternates between 0 and x/2 an infinite number of times. But you can see that as x gets close to 0 the value of x/2 also gets close to zero. So if I challenge you by asking what value of x will guarrantee that f(x) is less than, say, 0.01, you can see that this is true if x<= 0.02. For all values less than 0.02 the value of f(x) is less than 0.01. Thus this function passes the "epsilon-delta" test - for any epsilon you can name (0.01 in this exampe) it's possible to find a delta below which f(x)< epsilon.

    Now think about the second problem - as x gets small and approaches zero what value - if any - does f(x) approach?
    Last edited by ebaines; August 7th 2013 at 06:03 AM.
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  3. #3
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    Re: Limits

    Quote Originally Posted by Kristen111111111111111111 View Post
    If f(x)=x/2 when x belongs to rational numbers
    =0 when x belongs to irrational numbers
    Can limit x reaches 0 exist?
    Yes, it should be clear that the limit is 0. Given any epsilon> 0, take delta to be a number less than 2epsilon. If |x- 0|= |x|< delta, then either x is irrational, so that f(x)= 0< epsilon or x is rational so that |f(x)|= |x|/2< delta/2< epsilon. In either case, |f(x)- 0|= |f(x)|< epsilon.

    And If g(x)=0 when x belongs to rational numbers
    =x+1 when x belongs to irrational numbers
    Can limit x reaches 0 exist?
    Suppose x is an irrational number, very close to 0. What can you say about f(x)? Suppose x is a rational number, very close to 0?

    Can someone please explain about these?
    Last edited by HallsofIvy; August 7th 2013 at 06:05 AM.
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    Re: Limits

    Thank you so much. From your explanations I got that limit x reaches 0 exists for f(x) and limit x reaches 0 does not exist for g(x).
    And just to clarify I need to know whether a limit exist for a function defined only on the set of rational numbers?
    For example f(x) = x/2 , for all x belongs to rational numbers , does the limit x reaches 0 exist?
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  5. #5
    MHF Contributor ebaines's Avatar
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    Re: Limits

    Quote Originally Posted by Kristen111111111111111111 View Post
    Thank you so much. From your explanations I got that limit x reaches 0 exists for f(x) and limit x reaches 0 does not exist for g(x).
    And just to clarify I need to know whether a limit exist for a function defined only on the set of rational numbers?
    For example f(x) = x/2 , for all x belongs to rational numbers , does the limit x reaches 0 exist?
    Depends how you define the value off(x) of x = irrational numbers. If you say f(x) is undefined for irrational numbers then no, there is no value for the limit.
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  6. #6
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    Re: Limits

    Yeah got that. Thank you
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