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Math Help - Intedeterminate froms and L'Hospital's rule

  1. #1
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    Intedeterminate froms and L'Hospital's rule

    I cant seem to get these few....

    any help??

    1. lim ((x)/(x-1) - (1)/(ln x))
    x-> 1

    2. Lim (sqrt(x^2-x) - x)
    x-> INFINITY

    3. Lim (cot x - (1/x))
    x -> 0
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  2. #2
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    Quote Originally Posted by skabani View Post
    I cant seem to get these few....

    any help??

    1. lim ((x)/(x-1) - (1)/(ln x))
    x-> 1

    2. Lim (sqrt(x^2-x) - x)
    x-> INFINITY

    3. Lim (cot x - (1/x))
    x -> 0
    2. Note:

    \sqrt{x^2 - x} - x = \frac{(\sqrt{x^2 - x} - x) \, (\sqrt{x^2 - x} + x)}{(\sqrt{x^2 - x} + x)} = \frac{(x^2 - x) - x^2}{\sqrt{x^2 - x} + x} = \frac{-x}{\sqrt{x^2 - x} + x}.
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  3. #3
    o_O
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    1. \lim_{x \to 1} \left(\frac{x}{x-1} - \frac{1}{\ln x}\right)
    = \lim_{x \to 1} \left(\frac{x\ln x - (x - 1)}{\ln x(x - 1)}\right)
    = \lim_{x \to 1} \left(\frac{x\ln x - x + 1}{\ln x(x-1)}\right) = \left[ \frac{0}{0}\right]

    2 \lim_{x \to \infty} \left[ \left( \sqrt{x^{2} - x}-x\right) \cdot \left(\frac{\sqrt{x^{2} - x} + x}{\sqrt{x^{2} - x} + x}\right) \right]

    Should be pretty straight forward.

    3. \lim_{x \to 0} \left( \cot x - \frac{1}{x} \right)
    = \lim_{x \to 0} \left(\frac{1}{\tan x} - \frac{1}{x}\right)
    = \lim_{x \to 0} \left(\frac{x - \tan x}{x\tan x}\right) = \left[ \frac{0}{0} \right]
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  4. #4
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    I think the syntax for number 1 got mixed up..do you mind re-posting it??

    Thanks so much!!
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  5. #5
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    nvm..thanks
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  6. #6
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    Quote Originally Posted by o_O View Post
    1. \lim_{x \to 1} \left(\frac{x}{x-1} - \frac{1}{\ln x}\right)
    = \lim_{x \to 1} \left(\frac{x\ln x - (x - 1)}{\ln x(x - 1)}\right)
    = \lim_{x \to 1} \left(\frac{x\ln x - x + 1}{\ln x(x-1)}\right) = \left[ \frac{0}{0}\right]

    2 \lim_{x \to \infty} \left[ \left( \sqrt{x^{2} - x}-x\right) \cdot \left(\frac{\sqrt{x^{2} - x} + x}{\sqrt{x^{2} - x} + x}\right) \right]

    Should be pretty straight forward.

    3. \lim_{x \to 0} \left( \cot x - \frac{1}{x} \right)
    = \lim_{x \to 0} \left(\frac{1}{\tan x} - \frac{1}{x}\right)
    = \lim_{x \to 0} \left(\frac{x - \tan x}{x\tan x}\right) = \left[ \frac{0}{0} \right]
    i have the answer to 1 and its (1/2), not 0...
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  7. #7
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    Quote Originally Posted by skabani View Post
    i have the answer to 1 and its (1/2), not 0...
    o_O never said the answer to number 1 is zero, (s)he said it was the indeterminant form 0/0.

    -Dan
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  8. #8
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    Quote Originally Posted by o_O View Post
    [snip]
    3. \lim_{x \to 0} \left( \cot x - \frac{1}{x} \right)
    = \lim_{x \to 0} \left(\frac{1}{\tan x} - \frac{1}{x}\right)
    = \lim_{x \to 0} \left(\frac{x - \tan x}{x\tan x}\right) = \left[ \frac{0}{0} \right]
    For 3., instead of running to the l'hopital (several times in fact) - help! help! it's an emergency - you could instead substitute the series expansion for tan x around x = 0. Then you have:


    \lim_{x \to 0} \left[\, \frac{x - (x + x^3/3 + 2x^5/15 + ....)}{x (x + x^3/3 + 2x^5/15 + ....)}\, \right] = \lim_{x \to 0} \left[\, \frac{-x^3/3 - 2x^5/15 - ....)}{x^2 + x^4/3 + 2x^6/15 + ....)}\, \right]


    \lim_{x \to 0} \left[\, \frac{-x/3 - 2x^2/15 - ....)}{1 + x^2/3 + 2x^4/15 + ....)}\, \right] = 0.
    Last edited by mr fantastic; April 4th 2008 at 08:00 PM.
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  9. #9
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    Quote Originally Posted by o_O View Post
    1. \lim_{x \to 1} \left(\frac{x}{x-1} - \frac{1}{\ln x}\right)
    = \lim_{x \to 1} \left(\frac{x\ln x - (x - 1)}{\ln x(x - 1)}\right)
    = \lim_{x \to 1} \left(\frac{x\ln x - x + 1}{\ln x(x-1)}\right) = \left[ \frac{0}{0}\right]

    [snip]
    Again, instead of running to the l'Hopital, you could substitute the series expansion for ln x around x = 1:

    \ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - .....

    Then:


    \frac{x \ln x - (x - 1)}{(x-1) \ln x} = \frac{x \left( (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - .... \right) - (x - 1)}{(x-1) \left( (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - .... \right)}


    = \frac{x \left(1 - \frac{(x-1)}{2} + \frac{(x-1)^2}{3} - .... \right) - 1}{(x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - ....}


    = \frac{(x - 1) - \frac{x(x-1)}{2} + \frac{x(x-1)^2}{3} - ....}{(x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - ....}


     = \frac{1 - \frac{x}{2} + \frac{x(x-1)}{3} - ....}{1 - \frac{(x-1)}{2} + \frac{(x-1)^2}{3} - ....}


    and it's pretty clear what happens as x \rightarrow 1 ......
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