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Math Help - Uniform continuity help

  1. #1
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    Uniform continuity help

    I have to detemine if these functions are uniformly continuous on the given intervals

    1) f(x) = ln X on (0,1)
    my solution :
    Let ε >0 , we can let δ=ε
    by the mean value theorm says there is c belongs to (0,1) such that
    ln(X)-ln(Y)\ x-y =1\c <= 1
    thus, whenever x-y < δ
    So , we have ln(x) - ln(y)<= x-y< δ=ε
    so the function is uniformly continuous on the interval .

    Is my solution right ???


    2)x\tan x on the interval (- infinity, infinity)
    my solution :
    tan-1(x) = cotan(x), it's not uniformly continuous and becomes infinitely large at the points pik, k = 0,1,2,3, x/tan(x) and not uniformly continuous at given interval.

    I do not know how can i prove this by the definition . can anybody help me???


    3)f(x)= cos(lnx) on the interval (0,1)
    my solution:
    f(x) is continuous on the interval (0,1)
    for any two points from the interval (0,1) we can write the following expression:
    |f(x1) - f(x2)| = |cos(ln(x1) - cos(ln(x2)| = |- 1/2 sin (lnx1 + lnx2)/2 sin (lnx1 - lnx2)/2 | = = 1/2 |sin(ln(x1x2)/2) sin (ln(x1/x2)/2)| <= 1/2, so the assumption of uniformly continuity is performed at this interval.

    Is my solution right ?


    4)f(x)=x2 tan-1x [0, infinity)
    My solution :
    I think it is the same on the first question
    it's not continuous and becomes infinitely large at the points pik, k = 0,1,2,3, x/tan(x) and not continuous at given interval.

    5)ex
    on the interval (0,infinity)
    my solution :
    it is not uniformly continuous function , but i do not know how i can prove this by the definition



    Please can anybody help me to solve these questions ???
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  2. #2
    Super Member girdav's Avatar
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    Quote Originally Posted by mariama View Post
    I have to detemine if these functions are uniformly continuous on the given intervals

    1) f(x) = ln X on (0,1)
    my solution :
    Let ε >0 , we can let δ=ε
    by the mean value theorm says there is c belongs to (0,1) such that
    ln(X)-ln(Y)\ x-y =1\c <= 1
    thus, whenever x-y < δ
    So , we have ln(x) - ln(y)<= x-y< δ=ε
    so the function is uniformly continuous on the interval .

    Is my solution right ???
    No because you can't say that \frac 1c \leq 1 whereas 0<c\leq 1.
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  3. #3
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    Quote Originally Posted by girdav View Post
    No because you can't say that \frac 1c \leq 1 whereas 0<c\leq 1.
    So , how can i do it ?
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  4. #4
    Super Member girdav's Avatar
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    You can't show the function is uniformly continuous because it is not the case. Let x_n=e^{-n} and y_n =2e^{-n}. We have |x_n-y_n| =e^{-n} and what about |\ln x_n-\ln y_n| ?
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  5. #5
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    Quote Originally Posted by girdav View Post
    You can't show the function is uniformly continuous because it is not the case. Let x_n=e^{-n} and y_n =2e^{-n}. We have |x_n-y_n| =e^{-n} and what about |\ln x_n-\ln y_n| ?
    So, we have to approve that Lim | f(x+h)-f(x) |= infinty when x goes to a , and x goes to b
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  6. #6
    Super Member girdav's Avatar
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    We don't now what a and b are.
    Why don't you try what I proposed?
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  7. #7
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    Quote Originally Posted by girdav View Post
    We don't now what a and b are.
    Why don't you try what I proposed?
    I mean that , if this function is not uniformly continuous , so we have to prove that, not by defenition.
    We have to approve that Lim when x goes to a and Lim when x goes to b for f(x+h)-f(x) = inifinty . where a, b the end points of the interval (a,b)
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