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Math Help - Continuity at a point

  1. #1
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    Continuity at a point

    Set f: (0,1) --> R by:

    f(x) = (1/sqrt(x))-sqrt((x+1)/(x))

    Can one definite f(0) to make f continuous at ? Explain.


    I believe you can set the limit as x approaches 0 equal to f(0) in order to make it continuous.

    That is

    the limit as f(x) as x->0 is 0, so you could set f(0)=0 to make this happen?

    Is this correct?
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  2. #2
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    Quote Originally Posted by zhupolongjoe View Post
    Set f: (0,1) --> R by:

    f(x) = (1/sqrt(x))-sqrt((x+1)/(x))

    Can one definite f(0) to make f continuous at ? Explain.


    I believe you can set the limit as x approaches 0 equal to f(0) in order to make it continuous.

    That is

    the limit as f(x) as x->0 is 0, so you could set f(0)=0 to make this happen?

    Is this correct?

    Yes.

    Tonio
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  3. #3
    MHF Contributor chisigma's Avatar
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    With simple passages is...

    f(x)= \frac{1-\sqrt{1+x}}{\sqrt{x}} = \frac{\sqrt{x}}{1+\sqrt{1+x}} (1)

    All right!... but necessary condition for f(x) to be continous in x=0 is...

    \lim_{x \rightarrow 0+} f(x) = \lim_{x \rightarrow 0-} f(x) = f(0) (2)

    Is that satisfied for (1)? ...

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by chisigma View Post
    With simple passages is...

    f(x)= \frac{1-\sqrt{1+x}}{\sqrt{x}} = \frac{\sqrt{x}}{1+\sqrt{1+x}} (1)
    Actually, it is -\frac{\sqrt{x}}{1+ \sqrt{1+x}}

    All right!... but necessary condition for f(x) to be continous in x=0 is...

    \lim_{x \rightarrow 0+} f(x) = \lim_{x \rightarrow 0-} f(x) = f(0) (2)

    Is that satisfied for (1)? ...

    Kind regards

    \chi \sigma
    It is "continuous from the right" if you set f(0)= 0.

    And many text books use the term "continuous" to mean that if \{x_n\} is a sequence of points in the domain of f converging to a, then \lim_{n\to\infty} f(x_n)= f(x).
    Last edited by mr fantastic; October 24th 2009 at 05:51 AM. Reason: Added close quote tag to first quote.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Actually, it is -\frac{\sqrt{x}}{1+ \sqrt{1+x}}

    In the complex z plain the function...

    f(z)= \frac{\sqrt{z}}{1+\sqrt{1+z}} (1)

    ... is a multivalued function and in particular it has four independent branches connecting two branch points: z=0 and z=-1. When z=x ,  x real the four branches can be represented as...

    f(x)= \frac{\pm \sqrt{x}}{1 \pm \sqrt{1+x}} (2)

    ... and I reported for semplicity sake one of them...

    Quote Originally Posted by HallsofIvy View Post

    ... it is 'continuous from the right' if you set f(0)= 0...

    ... many text books use the term 'continuous' to mean that if \{x_n\} is a sequence of points in the domain of f converging to a, then \lim_{n\to\infty} f(x_n)= f(x).
    ... You say 'many text books use' and that means automatically that 'many text books don't' anf that is enough to make your point of view a little questionable...

    Kind regards

    \chi \sigma
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  6. #6
    MHF Contributor chisigma's Avatar
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    A more interesting consideration regarding the multivalued function...

    f(x) = \frac{\pm \sqrt{x}}{1 \pm \sqrt{1+x}} (1)

    ... is that if we choose the sign '-' in denominator is...

    \lim_{x \rightarrow 0+} f(x) = \pm \infty \ne 0 (2)

    ... fancy that! ...

    Kind regards

    \chi \sigma
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