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Math Help - use LIMIT definition

  1. #1
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    use LIMIT definition

    i need help

    with the use of the limit definition how can i prove this

    lim 1 = 1 c is not zero
    x-c x c

    also
    If c>0, then lim square root of x = square root of c
    x-c
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  2. #2
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    Quote Originally Posted by loop View Post
    i need help

    with the use of the limit definition how can i prove this

    lim 1 = 1 c is not zero
    x-c x c
    please format your question better. i don't know what you are asking

    also
    If c>0, then lim square root of x = square root of c
    x-c
    do you know what the definition says?

    note that |\sqrt x - \sqrt c| = \left| (\sqrt x - \sqrt c) \cdot \frac {\sqrt x + \sqrt c}{\sqrt x + \sqrt c} \right| = \frac {|x - c|}{\sqrt x + \sqrt c} \le \frac {|x - c|}{\sqrt c}

    for c \ne 0


    now can you fill in the blanks?
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  3. #3
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    Quote Originally Posted by loop View Post
    i need help

    with the use of the limit definition how can i prove this

    lim 1 = 1 c is not zero Mr F says: I assume you mean {\color{red}\lim_{x \rightarrow c} \frac{1}{x} = \frac{1}{c}, ~ c \neq 0}
    x-c x c

    also
    If c>0, then lim square root of x = square root of c Mr F says: I assume you mean {\color{red}\lim_{x \rightarrow c} \sqrt{x} = \sqrt{c}, ~ c > 0}
    x-c
    ..
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  4. #4
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    yes that is what am trying to say.
    Last edited by mr fantastic; April 12th 2009 at 03:40 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by loop View Post
    sorry about the post. and yes that is what am trying to say.
    for the first, we want \left| \frac 1x - \frac 1c \right| = \left| \frac {c - x}{xc} \right| < \epsilon for any \epsilon > 0, provided |x - c| < \delta for some \delta > 0.

    can you fill in the blanks?
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  6. #6
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    Quote Originally Posted by loop View Post
    i need help

    with the use of the limit definition how can i prove this

    lim 1 = 1 c is not zero
    x-c x c

    also
    If c>0, then lim square root of x = square root of c
    x-c
    The first one has also been asked and replied to here: http://www.mathhelpforum.com/math-he...ove-limit.html
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  7. #7
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    Quote Originally Posted by loop View Post
    [snip]
    If c>0, then lim square root of x = square root of c
    x-c
    | f(x) - l | = | \sqrt{x} - \sqrt{c} | = \left| \frac{(\sqrt{x} - \sqrt{c} ) (\sqrt{x} + \sqrt{c} )}{\sqrt{x} + \sqrt{c} } \right| = \left| \frac{x - c}{\sqrt{x} + \sqrt{c} }\right| = \frac{1}{\sqrt{x} + \sqrt{c}} \, | x - c |.

    Only values of x in the neighbourhood of c are of interest so impose the restriction 0 < x < 2c \Rightarrow 0 < | x - c| < c, say.

    Under this restriction, \frac{1}{\sqrt{x} + \sqrt{c}} < \frac{1}{\sqrt{c}}. Therefore:

    | f(x) - l | = \frac{1}{\sqrt{x} + \sqrt{c}} \, | x - c | < \frac{1}{\sqrt{c}} | x - c |.

    Note that  | f(x) - l | < \frac{1}{\sqrt{c}} | x - c | < \epsilon \Rightarrow | x - c| < \sqrt{c} \, \epsilon.

    So for all \epsilon > 0 there exists \delta (\epsilon) = \min\{ c, \, \sqrt{c} \, \epsilon\} such that 0 < | x - c | < \delta \Rightarrow | \sqrt{x} - \sqrt{c} | < \epsilon.
    Last edited by mr fantastic; April 12th 2009 at 11:36 PM. Reason: Added a small technical detail
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