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Math Help - Evaluating limits.

  1. #1
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    Evaluating limits.

    find the given limit as x approaches 0.
    ([1/(x+1)] /x) (-1/x)

    please help I'm confused, is there a way to find this algebraically?

    thanks,
    zirr
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  2. #2
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    Quote Originally Posted by Zirrick View Post
    find the given limit as x approaches 0.
    ([1/(x+1)] /x) (-1/x)

    please help I'm confused, is there a way to find this algebraically?

    thanks,
    zirr
    Is this what you are tying to solve:

    \displaystyle\lim_{x\to 0}\left[\frac{\frac{1}{x+1}}{x}\cdot\frac{-1}{x}\right]
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    Is this what you are tying to solve:
    \displaystyle\lim_{x\to 0}\left[\frac{\frac{1}{x+1}}{x}\cdot\frac{-1}{x}\right]
    If  \displaystyle\left[\frac{\frac{1}{x+1}}{x}\cdot\frac{-1}{x}\right] is correct then note that is equal to
     \displaystyle\left[{\frac{x}{x+1}\cdot\frac{-1}{x}\right]=\frac{-1}{x+1}
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  4. #4
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    Quote Originally Posted by dwsmith View Post
    Is this what you are tying to solve:

    \displaystyle\lim_{x\to 0}\left[\frac{\frac{1}{x+1}}{x}\cdot\frac{-1}{x}\right]
    um, not exactly i think i wrote it wrong. maybe like this?
    [(1/x+1)-1/x]
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  5. #5
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    Quote Originally Posted by Zirrick View Post
    um, not exactly i think i wrote it wrong. maybe like this?
    [(1/x+1)-1/x]
    \displaystyle\frac{1}{x+1}-\frac{1}{x}=\frac{x-x-1}{x(x+1)}=\frac{-1}{x(x+1)}

    \displaystyle\lim_{x\to 0}\frac{-1}{x(x+1)}

    Now, is this correct?
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  6. #6
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    I'm sorry I'm not to familiar with how exactly to write this.

    [(1/(x+1))-1]/x

    thanks for trying to help.
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  7. #7
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    Quote Originally Posted by Zirrick View Post
    [(1/(x+1))-1]/x

    thanks for trying to help.
    \displaystyle\frac{\frac{1}{x+1}-1}{x}\text{?}

    This one makes sense since it utilizes L'Hopitals Rule.
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  8. #8
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    Yes! thats the exact way its written!
    Thanks, what are the possible ways to find the limit? as x approaches 0.
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  9. #9
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    Quote Originally Posted by Zirrick View Post
    Yes! thats the exact way its written!
    Thanks, what are the possible ways to find the limit? as x approaches 0.
    \displaystyle\lim_{x\to 0}\left[\frac{\frac{1}{x+1}-1}{x}\right]=\frac{0}{0}

    \displaystyle\lim_{x\to 0}\frac{f'(x)}{g'(x)}\Rightarrow\lim_{x\to 0}\frac{\frac{-1}{(x+1)^2}}{1}=-1
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  10. #10
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    can you explain how to use L'Hopital's rule?
    Last edited by Zirrick; January 30th 2011 at 07:08 PM. Reason: re-write
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  11. #11
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    Quote Originally Posted by Zirrick View Post
    is there other way to do this?
    or is this the only way?
    \displaystyle\frac{\frac{1}{x+1}-1}{x}=\frac{\frac{1-x-1}{x+1}}{x}=\frac{\frac{-x}{x+1}}{x}=\cdots

    After simplification, take the limit.
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  12. #12
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    Quote Originally Posted by Zirrick View Post
    can you explain how to use L'Hopital's rule?
    When you have an indeterminant form, you can take the derivative of both the numerator, denominator, and then check the limit.
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  13. #13
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    Quote Originally Posted by Plato View Post
    If  \displaystyle\left[\frac{\frac{1}{x+1}}{x}\cdot\frac{-1}{x}\right] is correct then note that is equal to
     \displaystyle\left[{\frac{x}{x+1}\cdot\frac{-1}{x}\right]=\frac{-1}{x+1}
    '
    No, it isn't. \frac{\frac{1}{x+1}}{x}= \frac{1}{x+1}\frac{1}{x}= \frac{1}{x(x+1)}
    NOT \frac{x}{x+1}. That would \frac{1}{\frac{x+1}{x}}.

    A very strange error for Plato to make. (What were you smoking? I want some!)
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  14. #14
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    Quote Originally Posted by dwsmith View Post
    When you have an indeterminant form, you can take the derivative of both the numerator, denominator, and then check the limit.
    If f(x)-> 0 and g(x)-> 0 as x->a, then [tex]\lim_{x\to a}\frac{f(x)}{g(x)}= \lim_{x\to a}\frac{f'(x)}{g'(x)}[/itex]
    and you can extend to \frac{\infty}{\infty} and other "indeterminant forms".

    Note that numerator and denominator are differentiated separately, you are not differntiating the fraction using the quotient law.
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