Question

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

Answer 1

l'hopital for this one

Answer 2

take derivative top and bottom get
\[\lim_{x\rightarrow 1}\frac{x^3-3x^2+3x}{-3(1-x)^2}\]

Answer 3

wow that was wrong!

Answer 4

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
\[ \lim_{x->1} \frac{ \ln(3x-x^2+x^3)-0}{-3(1-x)^2}\]

Answer 5

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

Answer 6

i forgot the log, sorry

Answer 7

Um the differentiation of the numerator is valid due to FTC 2 right?

Answer 8

yup. derivative of integral is integrand

Answer 9

now on differentiating again
\[ \lim_{x->1} \frac{\frac{3x^2-6x+3}{x^3-3x^2+3x}}{-6(1-x)}\]

Answer 10

gonna hae to do it again though

Answer 11

ash has it

Answer 12

@ash
x\rightarrow 1
\[x\rightarrow 1\]

Answer 13

Thanks sat :)

Answer 14

sat: x \to 1 also work ;)

Answer 15

really?

Answer 16

\[x\to\]

Answer 17

YEsss :D

Answer 18

jeez and all this time. pluse i spell "rightarrow" wrong half the time

Answer 19

Let's differentiate it for the last time
\[\lim_{x\rightarrow1} \frac{\frac{(6x-6)(x^3-3x^2+3x)-(3x^2-6x+3)(3x^2-6x+3)}{(x^3-3x^2+3x)^2}}{+6}\]

Answer 20

\[\leftrightarrow\]

Answer 21

haha I was there too :P

Answer 22

i will post this in latexpractice

Answer 23

Sat, have you seen this http://openstudy.com/users/foolformath#/updates/4f2d5be5e4b0571e9cba67c0

Answer 24

wow!

Answer 25

If we substitute x=1 , numerator is zero and denominator is finite
so the answer is 0

Answer 26

ok thank you :)

Answer 27

welcome :)