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Math Help - Limit Question

  1. #1
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    Limit Question

    \lim_{x \to 0} \frac{(2x+1)^{1/3}-1}{x}
    Last edited by CaptainBlack; October 26th 2008 at 01:25 AM.
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  2. #2
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    Hello, treetheta!

    Edit: Made a blunder . . . will correct it now.

    We use: . (a - b)(a^2+ab+b^2) \:=\:a^3-b^3


    \lim_{x \to 0} \frac{(2x+1)^{\frac{1}{3}}-1}{x}

    Multiply top and bottom by: (2x+1)^{\frac{2}{3}} + (2x+1)^{\frac{1}{3}} + 1

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

    . . = \;\frac{2{\color{red}\rlap{/}}x}{{\color{red}\rlap{/}}x\left[(2x+1)^{\frac{2}{3}} + (2x+1)^{\frac{1}{3}} + 1\right]} \;=\;\frac{2}{(2x+1)^{\frac{2}{3}} + (2x+1)^{\frac{1}{3}} + 1}


    We have: . \lim_{x\to0}\left[\frac{2}{(2x+1)^{\frac{2}{3}} + (2x+1)^{\frac{1}{3}} + 1}\right] \;=\;\frac{2}{1^{\frac{2}{3}} + 1^{\frac{1}{3}} + 1} \;=\;\frac{2}{3}

    Last edited by Soroban; October 24th 2008 at 09:27 PM.
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  3. #3
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    how did this

    \lim_{x \to \o} \frac{(2x+1)^{1/3}-1}{x}

    go to

    \lim_{x \to \o} \frac{(2x+1)^{1/3}-1}{1}

    what happend to the x in the denominator

    ya, if u add the x, the answer becomes 2/3 thankssss ^^
    Last edited by treetheta; October 23rd 2008 at 09:42 PM.
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  4. #4
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    Is the limit supposed to be going to infinity or zero?
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  5. #5
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    It does, it's going to zero

    i put o instead of 0 sorry
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  6. #6
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    Quote Originally Posted by treetheta View Post
    \lim_{x \to 0} \frac{(2x+1)^{1/3}-1}{x}
    l'Hopital's Rule shows that the limit is equal to 2/3.

    Another way to do it is to substitute t = (2x + 1)^{1/3} \Rightarrow x = \frac{t^3 - 1}{2}. Then the limit becomes:

    \lim_{t \rightarrow 1} \frac{2(t-1)}{t^3 - 1}.

    Factorise the denominator, cancel the common factor of (t-1) in the numerator and denominator and then take the limit.
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  7. #7
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by treetheta View Post
    \lim_{x \to 0} \frac{(2x+1)^{1/3}-1}{x}
    Yet another way : Let \begin{array}{rl}f:\mathbb{R} &\mapsto \mathbb{R}\\ x & \mapsto \sqrt[3]{2x+1}\end{array}. f is differentiable on \mathbb{R}-\left\{-\tfrac12\right\} and f'(x)=\frac{2/3}{(2x+1)^{2/3}}.


    \lim_{x \to 0} \frac{\sqrt[3]{2x+1}-1}{x}=\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}

    That's f'(0)...
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