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Math Help - find the value for which the function is continuous

  1. #1
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    find the value for which the function is continuous

    f(x)= ln x/(x-1), if 0<x not =1
    c , if x=1
    and what kind of discontinuity is present if c does not have this value?

    can someone help me with this problem?
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  2. #2
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    This is nearly impossible to read...

    Is it

    f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if }x \in (0, 1)\cup (1, \infty)\\c\textrm{ if }x=1\end{cases}

    or

    f(x) = \begin{cases}\ln{\left(\frac{x}{x - 1}\right)}\textrm{ if }x \in (0, 1)\cup (1, \infty)\\c\textrm{ if }x=1\end{cases}?
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  3. #3
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    f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if } 0<x not equal to 1 \\c\textrm{ if }x=1\end{cases}

    thats how it is i dont know how to do a not equal sign
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  4. #4
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    Quote Originally Posted by dat1611 View Post
    f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if } x>0 \, ; \, x \ne 1 \\c\textrm{ if }x=1\end{cases}
    fify
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  5. #5
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    Quote Originally Posted by dat1611 View Post
    f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if } 0<x not equal to 1 \\c\textrm{ if }x=1\end{cases}

    thats how it is i dont know how to do a not equal sign
    It is done by \neq

    [LaTeX ERROR: Convert failed]

    For the question at hand, you need to determine a value c such that the function is continuous. One way could be to just plug in 1 in the above definition with the natural log, but you cannot do that since you'd get something of the form 0/0. Hence you need to find the limit of [LaTeX ERROR: Convert failed] as x approaches 1. Or more succinctly, find

    [LaTeX ERROR: Convert failed]

    Once you find that, you can just set c equal to it. From the definitions of continuous functions, f(x) is continuous at point a if

    [LaTeX ERROR: Convert failed]
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  6. #6
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    ok can someone help with the actual problem
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  7. #7
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    Do you know L'Hopital's rule? If you do, just apply to the numerator and denominator and it ' should be pretty easy.
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  8. #8
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    Hello, dat1611!

    f(x)\;=\; \left\{ \begin{array}{cc}\dfrac{\ln x}{x-1}& x > 0,\;x \ne 1 \\ \\[-3mm]<br />
c  & x=1 \end{array} \right

    Find the value of c for which the function is continuous.

    To be continuous at x = 1, we want: . \displaystyle{\lim_{x\to1}\frac{\ln x}{x-1} \;=\;c  }


    The left side goes to \frac{0}{0} so we can apply L'Hopital.

    . \displaystyle{\lim_{x\to1}\frac{\frac{1}{x}}{1} \;=\;c \quad\Rightarrow\quad \lim_{x\to1}\frac{1}{x} \:=\:c


    Therefore: . c \:=\:1

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  9. #9
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    so
    $\displaystyle \lim\limit_{x \to 1}\frac{\ln{x}}{x - 1}$<br />
= 1
    so the answers 1?
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  10. #10
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    Yes!Although, it seems like you don't quite understand why. Do you need help with that?
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