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anonymous

  • 4 years ago

\[\lim_{x \rightarrow 1}\frac{\int\limits_{1}^{x}\ln (3y-3y ^{2}+y ^{3})dy}{(1-x)^{3}}\]

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  1. anonymous
    • 4 years ago
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    l'hopital for this one

  2. anonymous
    • 4 years ago
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    take derivative top and bottom get \[\lim_{x\rightarrow 1}\frac{x^3-3x^2+3x}{-3(1-x)^2}\]

  3. anonymous
    • 4 years ago
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    wow that was wrong!

  4. ash2326
    • 4 years ago
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    start by differentiating numerator and denominator \[ \lim_{x->1} \frac{\frac{d}{dx}\int_{1}^{x} \ln(3y-y^2+y^3) dy}{\frac{d}{dx} (1-x)^3}\] we get \[ \lim_{x->1} \frac{ \ln(3x-x^2+x^3)-0}{-3(1-x)^2}\]

  5. anonymous
    • 4 years ago
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    \[\lim_{x\rightarrow 1}\frac{\ln(x^3-3x^2+3x)}{-3(1-x)^2}\]

  6. anonymous
    • 4 years ago
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    i forgot the log, sorry

  7. anonymous
    • 4 years ago
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    Um the differentiation of the numerator is valid due to FTC 2 right?

  8. anonymous
    • 4 years ago
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    yup. derivative of integral is integrand

  9. ash2326
    • 4 years ago
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    now on differentiating again \[ \lim_{x->1} \frac{\frac{3x^2-6x+3}{x^3-3x^2+3x}}{-6(1-x)}\]

  10. anonymous
    • 4 years ago
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    gonna hae to do it again though

  11. anonymous
    • 4 years ago
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    ash has it

  12. anonymous
    • 4 years ago
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    @ash x\rightarrow 1 \[x\rightarrow 1\]

  13. anonymous
    • 4 years ago
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    Thanks sat :)

  14. anonymous
    • 4 years ago
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    sat: x \to 1 also work ;)

  15. anonymous
    • 4 years ago
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    really?

  16. anonymous
    • 4 years ago
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    \[x\to\]

  17. anonymous
    • 4 years ago
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    YEsss :D

  18. anonymous
    • 4 years ago
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    jeez and all this time. pluse i spell "rightarrow" wrong half the time

  19. ash2326
    • 4 years ago
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    Let's differentiate it for the last time \[\lim_{x\rightarrow1} \frac{\frac{(6x-6)(x^3-3x^2+3x)-(3x^2-6x+3)(3x^2-6x+3)}{(x^3-3x^2+3x)^2}}{+6}\]

  20. anonymous
    • 4 years ago
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    \[\leftrightarrow\]

  21. anonymous
    • 4 years ago
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    haha I was there too :P

  22. anonymous
    • 4 years ago
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    i will post this in latexpractice

  23. anonymous
    • 4 years ago
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    Sat, have you seen this http://openstudy.com/users/foolformath#/updates/4f2d5be5e4b0571e9cba67c0

  24. anonymous
    • 4 years ago
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    wow!

  25. ash2326
    • 4 years ago
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    If we substitute x=1 , numerator is zero and denominator is finite so the answer is 0

  26. anonymous
    • 4 years ago
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    ok thank you :)

  27. ash2326
    • 4 years ago
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    welcome :)

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