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  1. #1 limit question 
    Moderator Moderator AlexP's Avatar
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    i'm currently reading on my own about limits, and the only "x approaches" exposure i've had before is "as x approaches infinity" in connection with Euler's Law. Obviously, x approaches, but never actually reaches, infinity. However, if you have, for example, "x--->2", does x actually reach 2, or does it just grow ever-closer to it without ever actually reaching it?


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  3. #2  
    Forum Professor river_rat's Avatar
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    It may help to give the actual definition and to unpack it.

    Definition: A function f(x) from R to R is said to approach L as x approaches t iff for every epsilon > 0 there exists a delta > 0 such that if |x-t| < delta then |f(x) - L| < epsilon.

    Now the trick here is that delta is greater then zero, so x is never equal to t and so x never reaches t (or 2 in your question).


    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  4. #3  
    Moderator Moderator AlexP's Avatar
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    that's the answer i was looking for, thanks
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  5. #4  
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    well, x can be equal to t because delta is just an upper bound on |x-t|.
    where are we? what are we?
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  6. #5  
    Forum Professor river_rat's Avatar
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    Oops - your right, here is the corrected definition

    Definition: A function f(x) from R to R is said to approach L as x approaches t iff for every epsilon > 0 there exists a delta > 0 such that if 0 < |x-t| < delta then |f(x) - L| < epsilon.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  7. #6  
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    consider this case

    limit (x-2)/(x-4)
    x-->2

    in this case you can assume that x takes the value 2 as there is no discontinuity or abberation here

    the answer for the above limit is 0

    but in this case

    lim (x-2)/(x-2) the denominator tends to 0 as x keeps moving towards 2
    x-->2

    when x=2, the expression becomes 0/0 which is indeterminate. so we assume that x only approches 2 and see what becomes of the expression. as x tends to 2 the value of the function is the value to which the expression points to

    in this case the expression tends to 1 as x tends to two and the expression is not defined when x equals two
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  8. #7  
    Moderator Moderator AlexP's Avatar
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    Quote Originally Posted by hypercube
    consider this case

    limit (x-2)/(x-4)
    x-->2

    in this case you can assume that x takes the value 2 as there is no discontinuity or abberation here

    the answer for the above limit is 0

    but in this case

    lim (x-2)/(x-2) the denominator tends to 0 as x keeps moving towards 2
    x-->2

    when x=2, the expression becomes 0/0 which is indeterminate. so we assume that x only approches 2 and see what becomes of the expression. as x tends to 2 the value of the function is the value to which the expression points to

    in this case the expression tends to 1 as x tends to two and the expression is not defined when x equals two
    makes sense, thanks for taking the time to reply.

    gotta love those hypercubes...
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  9. #8  
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    any time chemy any time :-D
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  10. #9 hai chem 
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    iam sravan......just recently joined this forum



    ur question totally depends on the function ur dealing with

    case 1:

    if the function is discontinuous at X=2 then X actually will be in the smallest neigbourhood around 2 then u have the L'Hospitals Rule for caluculating the value of function at X=2

    case 2:

    if the function is continuous at X=2 the u can just substitute the value of X in the function and get the value of the function directly.........here X actually approaches 2
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