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Thread: limit question..

  1. #1
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    limit question..

    the question and how i tried to solve it
    in this link:
    http://img368.imageshack.us/img368/6763/99124375pc4.gif

    i dont know how
    ??
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  2. #2
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    $\displaystyle \lim_{t \rightarrow 0^{-}}\sqrt{\frac{1}{t^{3}}}\left(\sqrt{\frac{2t+1}{t} }-2\sqrt{\frac{t+1}{t}}+\sqrt{\frac{1}{t}} \right) \Rightarrow $ undefined

    $\displaystyle
    \lim_{t \rightarrow 0^{+}}\sqrt{\frac{1}{t^{3}}}\left(\sqrt{\frac{2t+1 }{t}}-2\sqrt{\frac{t+1}{t}}+\sqrt{\frac{1}{t}} \right)\Rightarrow -\frac{1}{4}
    $

    As you can see this limit is only defined when x approach 0 from the right.
    I concluded that the limit was -1/4 by looking at the graph of $\displaystyle f(x) = \sqrt{x^{3}}\left(\sqrt{2x+1}- 2\sqrt{x+1} +\sqrt{x} \right) $ as x was approaching infinity.
    Last edited by fonso_gfx; Dec 19th 2008 at 09:58 AM.
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  3. #3
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    how did you transform
    $\displaystyle

    f(x) = \sqrt{x^{3}}\left(\sqrt{2x+1}- 2\sqrt{x+1} +\sqrt{x} \right)
    $

    into

    $\displaystyle

    \lim_{x \rightarrow
    0^{+}}\sqrt{\frac{1}{x^{3}}}\left(\frac{2x+1}{x}-2\sqrt{\frac{x+1}{x}}+\sqrt{\frac{1}{x}}
    \right)\Rightarrow -\frac{1}{4}
    $
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    how did you transform
    $\displaystyle

    f(x) = \sqrt{x^{3}}\left(\sqrt{2x+1}- 2\sqrt{x+1} +\sqrt{x} \right)
    $

    into

    $\displaystyle

    \lim_{x \rightarrow
    0^{+}}\sqrt{\frac{1}{x^{3}}}\left(\frac{2x+1}{x}-2\sqrt{\frac{x+1}{x}}+\sqrt{\frac{1}{x}}
    \right)\Rightarrow -\frac{1}{4}
    $

    Well, i got from your problem (where you let x = 1/t)
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  5. #5
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    how did you get the 1/4 result in here??

    $\displaystyle
    \lim_{t \rightarrow 0^{+}}\sqrt{\frac{1}{t^{3}}}\left(\sqrt{\frac{2t+1 }{t}}-2\sqrt{\frac{t+1}{t}}+\sqrt{\frac{1}{t}} \right)\Rightarrow -\frac{1}{4}
    $
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    how did you get the 1/4 result in here??

    $\displaystyle
    \lim_{t \rightarrow 0^{+}}\sqrt{\frac{1}{t^{3}}}\left(\sqrt{\frac{2t+1 }{t}}-2\sqrt{\frac{t+1}{t}}+\sqrt{\frac{1}{t}} \right)\Rightarrow -\frac{1}{4}
    $

    I looked at the table of values for f(x) in my calculator.
    f(0.00001) is approx -0.250136 which is about -1/4
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  7. #7
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    i cant solve like that
    i need mathematical way
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    how did you get the 1/4 result in here??

    $\displaystyle
    \lim_{t \rightarrow 0^{+}}\sqrt{\frac{1}{t^{3}}}\left(\sqrt{\frac{2t+1 }{t}}-2\sqrt{\frac{t+1}{t}}+\sqrt{\frac{1}{t}} \right)\Rightarrow -\frac{1}{4}
    $
    Quote Originally Posted by transgalactic View Post
    i cant solve like that
    i need mathematical way
    Supposing this guy up here has the correct limit you have

    $\displaystyle \lim_{t \rightarrow 0^{+}}\sqrt{\frac{1}{t^{3}}}\left(\sqrt{\frac{2t+1 }{t}}-2\sqrt{\frac{t+1}{t}}+\sqrt{\frac{1}{t}} \right)=\lim_{t\to{0}}\frac{\sqrt{2t+1}-2\sqrt{t+1}+1}{t^2}$

    Apply the Binomial series for both to get

    $\displaystyle \begin{aligned}\lim_{t\to{0}}\frac{\left(1+t-\frac{t^2}{2}+\mathcal{O}(t^3)\right)-2\left(1+\frac{t}{2}-\frac{t^2}{8}+\mathcal{O}(t^3)\right)+1}{t^2}&=\li m_{t\to{0}}\frac{\frac{-t^2}{4}+\mathcal{O}(t^3)}{t^2}\\
    &=\frac{-1}{4}\end{aligned}$
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  9. #9
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    $\displaystyle \begin{aligned}\lim_{t\to{0}}\frac{\left(1+t-\frac{t^2}{2}+\mathcal{O}(t^3)\right)-2\left(1+\frac{t}{2}-\frac{t^2}{8}+\mathcal{O}(t^3)\right)+1}{t^2}&=\li m_{t\to{0}}\frac{\frac{-t^2}{4}+\mathcal{O}(t^3)}{t^2}\\
    &=\frac{-1}{4}\end{aligned}$[/quote]


    in the last part when you input t=0

    how you get -1/4
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    $\displaystyle \begin{aligned}\lim_{t\to{0}}\frac{\left(1+t-\frac{t^2}{2}+\mathcal{O}(t^3)\right)-2\left(1+\frac{t}{2}-\frac{t^2}{8}+\mathcal{O}(t^3)\right)+1}{t^2}&=\li m_{t\to{0}}\frac{\frac{-t^2}{4}+\mathcal{O}(t^3)}{t^2}\\
    &=\frac{-1}{4}\end{aligned}$

    in the last part when you input t=0

    how you get -1/4[/quote]
    Since as $\displaystyle t\to0$ we can ignore the terms denoted by $\displaystyle \mathcal{O}(t^3)$, this leaves us with $\displaystyle \frac{\frac{-t^2}{4}}{t^2}=\frac{-1}{4}$
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