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Thread: Limit

  1. #1
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    Limit

    How do i get 3/4?
    I've tried to group up the x and semplify them into the root... But this way the limit is equal to 0... Thx for your help. Have a nice dayLimit-screenshot_2019-01-05-01-56-54-479_com.wolfram.android.alpha.jpeg

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  2. #2
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    Re: Limit

    Quote Originally Posted by domy7997 View Post
    How do i get 3/4?
    I've tried to group up the x and semplify them into the root... But this way the limit is equal to 0... Thx for your help. Have a nice dayClick image for larger version. 

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    4/3*

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  3. #3
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    Re: Limit

    \begin{align}\lim_{x \to -\infty} \left(\sqrt[3]{\frac{5x^3-x^4}{x-1}} + x\right) &= \lim_{y \to +\infty} \left(\sqrt[3]{\frac{5y^3+y^4}{y+1}} - y\right) \\ &= \lim_{y \to +\infty} \left(\sqrt[3]{\frac{5y^3+y^4}{y+1}} - \sqrt[3]{\frac{y^3(y+1)}{y+1}} \right) \\ \end{align}
    Also $$a^3-b^3 = (a-b)(a^2 + ab + b^2)$$
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  4. #4
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    Re: Limit

    Quote Originally Posted by Archie View Post
    \begin{align}\lim_{x \to -\infty} \left(\sqrt[3]{\frac{5x^3-x^4}{x-1}} + x\right) &= \lim_{y \to +\infty} \left(\sqrt[3]{\frac{5y^3+y^4}{y+1}} - y\right) \\ &= \lim_{y \to +\infty} \left(\sqrt[3]{\frac{5y^3+y^4}{y+1}} - \sqrt[3]{\frac{y^3(y+1)}{y+1}} \right) \\ \end{align}
    Also $$a^3-b^3 = (a-b)(a^2 + ab + b^2)$$
    What did you do? How you know it is equal to 4/3? ^^

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  5. #5
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    Re: Limit

    Quote Originally Posted by domy7997 View Post
    What did you do? How you know it is equal to 4/3? ^^

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    Oh... False cube... Gonna give it a try. Ty!

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  6. #6
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    Re: Limit

    Or, setting $$t=\lim_{x \to -\infty} \left(\sqrt[3]{\frac{5x^3-x^4}{x-1}} + x\right)$$you can attack $e^t$ with l'H˘pital.
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  7. #7
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    Re: Limit

    Quote Originally Posted by Archie View Post
    \begin{align}\lim_{x \to -\infty} \left(\sqrt[3]{\frac{5x^3-x^4}{x-1}} + x\right) &= \lim_{y \to +\infty} \left(\sqrt[3]{\frac{5y^3+y^4}{y+1}} - y\right) \\ &= \lim_{y \to +\infty} \left(\sqrt[3]{\frac{5y^3+y^4}{y+1}} - \sqrt[3]{\frac{y^3(y+1)}{y+1}} \right) \\ \end{align}
    Also $$a^3-b^3 = (a-b)(a^2 + ab + b^2)$$
    If y=-x i think the substitution it's not complitely correct. I mean why 5x^3 does not become -5y^3? And why x-1 does not become -y-1?

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  8. #8
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    Re: Limit

    $\displaystyle \lim_{x \to -\infty} \left ( \sqrt[3]{ \frac{5x^3-x^4}{x-1} } + x \right )$

    Setting y = -x:
    $\displaystyle \lim_{(-y) \to -\infty} \left ( \sqrt[3]{ \frac{5(-y)^3-(-y)^4}{(-y)-1}} + (-y) \right )$

    There are a large number of substitutions here.
    1) In the limit $\displaystyle \lim_{-y \to -\infty}$ becomes $\displaystyle \lim_{y \to \infty}$

    2) $\displaystyle 5(-y)^3 = -5y^3$

    3) $\displaystyle (-y)^4 = y^4$

    4) $\displaystyle (-y) - 1 = -(y + 1)$

    Putting all these in we have
    $\displaystyle \lim_{y \to \infty} \left ( \sqrt[3]{ \frac{-5y^3-y^4}{-(y + 1)}} - y \right )$

    $\displaystyle = \lim_{y \to \infty} \left ( \sqrt[3]{ \frac{(-1)(5y^3+y^4)}{(-1)(y + 1)}} - y \right )$

    $\displaystyle = \lim_{y \to \infty} \left ( \sqrt[3]{ \frac{5y^3+y^4}{y + 1}} - y \right )$

    -Dan
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    Re: Limit

    Quote Originally Posted by domy7997 View Post
    it's not complitely correct.
    That's possible. I'm human after all.
    Quote Originally Posted by domy7997 View Post
    I mean why 5x^3 does not become -5y^3? And why x-1 does not become -y-1?
    I took a factor of $-1$ out of both denominator and numerator. $\tfrac{-1}{-1}=1$
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    Re: Limit

    Quote Originally Posted by Archie View Post
    That's possible. I'm human after all.

    I took a factor of $-1$ out of both denominator and numerator. $\tfrac{-1}{-1}=1$
    Lol... Actually sry. The fact is that i actually suck a bit. Btw i am still not able to continue it, even after the application of the false cube... Any suggestion?
    (You talked about hopital.How is it possible if i have no unique denominator?)
    Thx for your time mate!

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  11. #11
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    Re: Limit

    \begin{align}\lim_{ x \to -\infty } \left( \sqrt[3]{ \frac{ 5x^3-x^4 }{ x-1 } } + x \right) &= \lim_{ y \to +\infty } \left( \sqrt[3]{ \frac{ 5y^3+y^4 }{ y+1 } } - y \right) \\ &= \lim_{ y \to +\infty } \left( \sqrt[3]{ \frac{ 5y^3+y^4 }{ y+1 } } - \sqrt[3]{ \frac{ y^3(y+1) }{ y+1 } } \right) \\ &= \lim_{ y \to +\infty } \frac{ \left( \left( \frac{ 5y^3+y^4 }{ y+1 } \right)^\frac13 - \left( \frac{ y^3(y+1) }{ y+1 } \right)^\frac13 \right)\left( \left( \frac{ 5y^3+y^4 }{ y+1 } \right)^\frac23 + \left( \frac{ 5y^3+y^4 }{ y+1 } \right)^\frac13 \left( \frac{ y^3(y+1) }{ y+1 } \right)^\frac13 + \left( \frac{ y^3(y+1) }{ y+1 } \right)^\frac23 \right) }{ \left( \frac{ 5y^3+y^4 }{ y+1 } \right)^\frac23 + \left( \frac{ 5y^3+y^4 }{ y+1 } \right)^\frac13 \left( \frac{ y^3(y+1) }{ y+1 } \right)^\frac13 + \left( \frac{ y^3(y+1) }{ y+1 } \right)^\frac23 } \\ &= \lim_{ y \to +\infty } \frac{ \frac{ 5y^3+y^4 }{ y+1 } - \frac{ y^4+y^3 }{ y+1 } }{ \left( \frac{ 25y^6+10y^7+y^8 }{ (y+1)^2 } \right)^\frac13 + \left( \frac{ (5y^3+y^4)(y^4+y^3) }{ (y+1)^2 } \right)^\frac13 + \left( \frac{ (y^8+2y^7+y^6) }{ (y+1)^2 } \right)^\frac13 } \\ &= \lim_{ y \to +\infty } \frac{ \frac{ 4y^3 }{ y+1 } }{ \left( \frac{ (25y^6+10y^7+y^8)(y+1) }{ (y+1)^3 } \right)^\frac13 + \left( \frac{ (y^8 + 6y^7 + 5y^6)(y+1) }{ (y+1)^3 } \right)^\frac13 + \left( \frac{ (y^8+2y^7+y^6)(y+1) }{ (y+1)^3 } \right)^\frac13 } \\ &= \lim_{ y \to +\infty } \frac{ 4y^3 }{ \left( y^9+11y^8+35y^7+25y^6 \right)^\frac13 + \left( y^9+7y^8+11y^7+5y^6 \right)^\frac13 + \left( y^9+3y^8+3y^7+y^6 \right)^\frac13 } \\ &= \lim_{ y \to +\infty } \frac{ 4(y^9)^\frac13 }{ \left( y^9+11y^8+35y^7+25y^6 \right)^\frac13 + \left( y^9+7y^8+11y^7+5y^6 \right)^\frac13 + \left( y^9+3y^8+3y^7+y^6 \right)^\frac13 } \\ &= \lim_{ y \to +\infty } \frac{ 4(1)^\frac13 }{ \left( 1+11y^{ -1 }+35y^{ -2 }+25y^{ -3 } \right)^\frac13 + \left( 1+7y^{ -1 }+11y^{ -2 }+5y^{ -3 } \right)^\frac13 + \left( 1+3y^{ -1 }+3y^{ -2 }+y^{ -3 } \right)^\frac13 } \\ &= \frac{ 4(1)^\frac13 }{ \left( 1+0+0+0 \right)^\frac13 + \left( 1+0+0+0 \right)^\frac13 + \left( 1+0+0+0 \right)^\frac13 } \\ &= \frac{ 4(1)^\frac13 }{ 3\left( 1 \right)^\frac13 } \\ &= \frac43 \end{align}
    Last edited by Archie; Jan 4th 2019 at 07:22 PM.
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  12. #12
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    Re: Limit

    Quote Originally Posted by Archie View Post
    \begin{align}\lim_{ x \to -\infty } \left( \sqrt[3]{ \frac{ 5x^3-x^4 }{ x-1 } } + x \right) &= \lim_{ y \to +\infty } \left( \sqrt[3]{ \frac{ 5y^3+y^4 }{ y+1 } } - y \right) \\ &= \lim_{ y \to +\infty } \left( \sqrt[3]{ \frac{ 5y^3+y^4 }{ y+1 } } - \sqrt[3]{ \frac{ y^3(y+1) }{ y+1 } } \right) \\ &= \lim_{ y \to +\infty } \frac{ \left( \left( \frac{ 5y^3+y^4 }{ y+1 } \right)^\frac13 - \left( \frac{ y^3(y+1) }{ y+1 } \right)^\frac13 \right)\left( \left( \frac{ 5y^3+y^4 }{ y+1 } \right)^\frac23 + \left( \frac{ 5y^3+y^4 }{ y+1 } \right)^\frac13 \left( \frac{ y^3(y+1) }{ y+1 } \right)^\frac13 + \left( \frac{ y^3(y+1) }{ y+1 } \right)^\frac23 \right) }{ \left( \frac{ 5y^3+y^4 }{ y+1 } \right)^\frac23 + \left( \frac{ 5y^3+y^4 }{ y+1 } \right)^\frac13 \left( \frac{ y^3(y+1) }{ y+1 } \right)^\frac13 + \left( \frac{ y^3(y+1) }{ y+1 } \right)^\frac23 } \\ &= \lim_{ y \to +\infty } \frac{ \frac{ 5y^3+y^4 }{ y+1 } - \frac{ y^4+y^3 }{ y+1 } }{ \left( \frac{ 25y^6+10y^7+y^8 }{ (y+1)^2 } \right)^\frac13 + \left( \frac{ (5y^3+y^4)(y^4+y^3) }{ (y+1)^2 } \right)^\frac13 + \left( \frac{ (y^8+2y^7+y^6) }{ (y+1)^2 } \right)^\frac13 } \\ &= \lim_{ y \to +\infty } \frac{ \frac{ 4y^3 }{ y+1 } }{ \left( \frac{ (25y^6+10y^7+y^8)(y+1) }{ (y+1)^3 } \right)^\frac13 + \left( \frac{ (y^8 + 6y^7 + 5y^6)(y+1) }{ (y+1)^3 } \right)^\frac13 + \left( \frac{ (y^8+2y^7+y^6)(y+1) }{ (y+1)^3 } \right)^\frac13 } \\ &= \lim_{ y \to +\infty } \frac{ 4y^3 }{ \left( y^9+11y^8+35y^7+25y^6 \right)^\frac13 + \left( y^9+7y^8+11y^7+5y^6 \right)^\frac13 + \left( y^9+3y^8+3y^7+y^6 \right)^\frac13 } \\ &= \lim_{ y \to +\infty } \frac{ 4(y^9)^\frac13 }{ \left( y^9+11y^8+35y^7+25y^6 \right)^\frac13 + \left( y^9+7y^8+11y^7+5y^6 \right)^\frac13 + \left( y^9+3y^8+3y^7+y^6 \right)^\frac13 } \\ &= \lim_{ y \to +\infty } \frac{ 4(1)^\frac13 }{ \left( 1+11y^{ -1 }+35y^{ -2 }+25y^{ -3 } \right)^\frac13 + \left( 1+7y^{ -1 }+11y^{ -2 }+5y^{ -3 } \right)^\frac13 + \left( 1+3y^{ -1 }+3y^{ -2 }+y^{ -3 } \right)^\frac13 } \\ &= \frac{ 4(1)^\frac13 }{ \left( 1+0+0+0 \right)^\frac13 + \left( 1+0+0+0 \right)^\frac13 + \left( 1+0+0+0 \right)^\frac13 } \\ &= \frac{ 4(1)^\frac13 }{ 3\left( 1 \right)^\frac13 } \\ &= \frac43 \end{align}
    I swear a god... I wish i could read it x)
    Could you screenshot what you wrote? Maybe stamp resist it?... I'm in debt with you...^^Limit-screenshot_2019-01-05-05-27-02-955_com.android.chrome.jpeg

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  13. #13
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    Re: Limit

    Quote Originally Posted by domy7997 View Post
    (You talked about hopital.How is it possible if i have no unique denominator?)
    \begin{align}t &= \lim_{x \to -\infty} \left(\sqrt[3]{\frac{5x^3-x^4}{x-1}} + x\right) \\ e^t &= e^{\lim_{x \to -\infty} \left(\sqrt[3]{\frac{5x^3-x^4}{x-1}} + x\right)} \\ &= \lim_{x \to -\infty} e^{\left(\sqrt[3]{\frac{5x^3-x^4}{x-1}} + x\right)} \\ &= \lim_{x \to -\infty} e^{\sqrt[3]{\frac{5x^3-x^4}{x-1}}}e^{x} \\ &= \lim_{x \to -\infty} e^{\sqrt[3]{\frac{5x^3-x^4}{x-1}}}e^{-(-x)} \\ &= \lim_{x \to -\infty} \frac{e^{\sqrt[3]{\frac{5x^3-x^4}{x-1}}}}{e^{-x}} \\ \end{align}
    And now you can use l'H˘pital.
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    Re: Limit

    Quote Originally Posted by domy7997 View Post
    I swear a god... I wish i could read it x)
    Try refreshing the page.
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    Re: Limit

    Quote Originally Posted by Archie View Post
    Try refreshing the page.
    Nothing to do...

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