anonymous
  • anonymous
\[\lim_{x \rightarrow 1}\frac{\int\limits_{1}^{x}\ln (3y-3y ^{2}+y ^{3})dy}{(1-x)^{3}}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
l'hopital for this one
anonymous
  • anonymous
take derivative top and bottom get \[\lim_{x\rightarrow 1}\frac{x^3-3x^2+3x}{-3(1-x)^2}\]
anonymous
  • anonymous
wow that was wrong!

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More answers

ash2326
  • ash2326
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 \[ \lim_{x->1} \frac{ \ln(3x-x^2+x^3)-0}{-3(1-x)^2}\]
anonymous
  • anonymous
\[\lim_{x\rightarrow 1}\frac{\ln(x^3-3x^2+3x)}{-3(1-x)^2}\]
anonymous
  • anonymous
i forgot the log, sorry
anonymous
  • anonymous
Um the differentiation of the numerator is valid due to FTC 2 right?
anonymous
  • anonymous
yup. derivative of integral is integrand
ash2326
  • ash2326
now on differentiating again \[ \lim_{x->1} \frac{\frac{3x^2-6x+3}{x^3-3x^2+3x}}{-6(1-x)}\]
anonymous
  • anonymous
gonna hae to do it again though
anonymous
  • anonymous
ash has it
anonymous
  • anonymous
@ash x\rightarrow 1 \[x\rightarrow 1\]
anonymous
  • anonymous
Thanks sat :)
anonymous
  • anonymous
sat: x \to 1 also work ;)
anonymous
  • anonymous
really?
anonymous
  • anonymous
\[x\to\]
anonymous
  • anonymous
YEsss :D
anonymous
  • anonymous
jeez and all this time. pluse i spell "rightarrow" wrong half the time
ash2326
  • ash2326
Let's differentiate it for the last time \[\lim_{x\rightarrow1} \frac{\frac{(6x-6)(x^3-3x^2+3x)-(3x^2-6x+3)(3x^2-6x+3)}{(x^3-3x^2+3x)^2}}{+6}\]
anonymous
  • anonymous
\[\leftrightarrow\]
anonymous
  • anonymous
haha I was there too :P
anonymous
  • anonymous
i will post this in latexpractice
anonymous
  • anonymous
anonymous
  • anonymous
wow!
ash2326
  • ash2326
If we substitute x=1 , numerator is zero and denominator is finite so the answer is 0
anonymous
  • anonymous
ok thank you :)
ash2326
  • ash2326
welcome :)

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