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Thread: continious function

  1. #1
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    Question continious function

    f:[-1,1] -> R is defined as

    .......| 1/ln(x) , x not equal be 0
    f(x)={
    .......| 0, x=0

    how do i show this is continious???
    thankz
    Last edited by dopi; Nov 14th 2006 at 11:33 AM. Reason: error
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dopi View Post
    f:[-1,1] -> R is defined as

    .......| 1/ln(x) , x not equal be 0
    f(x)={
    .......| 0, x=0

    how do i show this is continious???
    thankz
    But ln(x) is only defined on (0, infinity)! Your function's domain is too large.

    -Dan
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  3. #3
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    ??

    Quote Originally Posted by topsquark View Post
    But ln(x) is only defined on (0, infinity)! Your function's domain is too large.

    -Dan
    no its not ...its defined on [-1, 1] isnt it?
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  4. #4
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    Maybe you mean:
    $\displaystyle f(x) = \left\{ {\begin{array}{lc}
    {\frac{1}{{\ln \left( {\left| x \right|} \right)}}} & {x \not= 0} \\
    0 & x = 0 \\
    \end{array}} \right.$

    Now it is continuous.
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  5. #5
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    Question

    Quote Originally Posted by Plato View Post
    Maybe you mean:
    $\displaystyle f(x) = \left\{ {\begin{array}{lc}
    {\frac{1}{{\ln \left( {\left| x \right|} \right)}}} & {x \not= 0} \\
    0 & x = 0 \\
    \end{array}} \right.$

    Now it is continuous.
    yea thats wat i meant...i forgot to put the modulus sign on...but how would i show that this is continious with f:[-1,1]
    thankz
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  6. #6
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    Still no good you should have mentioned the open interval $\displaystyle (-1,1)$


    Anyway, the way you show continuity is to show that,
    $\displaystyle \lim_{x\to 0}f(x)=f(0)=0$
    Since,
    $\displaystyle f(x)=\frac{1}{\ln|x|}$ open some open interval except possibly at $\displaystyle x=0$ we have,
    $\displaystyle \lim_{x\to 0}\frac{1}{\ln|x|}=0$
    Which is true if we consider the left-right limits,
    $\displaystyle \lim_{x\to 0^-}\frac{1}{\ln (-x)}=\lim_{x\to 0^+}\frac{1}{\ln x}$
    And by using L'Hopital's Rule.
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  7. #7
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    Quote Originally Posted by ThePerfectHacker View Post
    Still no good you should have mentioned the open interval $\displaystyle (-1,1)$


    Anyway, the way you show continuity is to show that,
    $\displaystyle \lim_{x\to 0}f(x)=f(0)=0$
    Since,
    $\displaystyle f(x)=\frac{1}{\ln|x|}$ open some open interval except possibly at $\displaystyle x=0$ we have,
    $\displaystyle \lim_{x\to 0}\frac{1}{\ln|x|}=0$
    Which is true if we consider the left-right limits,
    $\displaystyle \lim_{x\to 0^-}\frac{1}{\ln (-x)}=\lim_{x\to 0^+}\frac{1}{\ln x}$
    And by using L'Hopital's Rule.
    but its not open, its bounded [-1, 1] thats why i put square brackets...thankz for the help
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dopi View Post
    but its not open, its bounded [-1, 1] thats why i put square brackets...thankz for the help
    TPH was referring to the fact that $\displaystyle ln | \pm 1 | = 0$ so $\displaystyle \frac{1}{ln|x|}$ is undefined at $\displaystyle x = \pm 1$.

    -Dan
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