Problem: Limit of Definite Integral with ln(1-a), where a->1,a<1

Vect

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Initial Problem:
After taking the integral without its borders, and solving it with integration by parts:
https://www.wolframalpha.com/input/?i=limit+(ln(1-a)(a^2-1))/2+-(a(1+a/2))/2+as+a->1-
I don't know how to properly solve the limit. Maybe it's no biggie, but any help is appreciated.
I really don't know which is the actual question. It seems to be why:
21[lima1{(log(1a)(1a2)a(1+21a)}]=34 ?\displaystyle {\large{2^{ - 1}}\displaystyle\left[ {{{\lim }_{a \to {1^ - }}}\left\{ {\left( {\log (1 - a} \right)\left( {1 - {a^2}} \right) - a\left( {1 + {2^{\bf{ - 1}}}a} \right)} \right\}} \right]} = - \frac{3}{4}~?

That seems straightforward. NO?
 
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I really don't know which is the actual question. It seems to be why:
21[lima1{(log(1a)(1a2)a(121a)}]=34 ?\displaystyle {\large{2^{ - 1}}\displaystyle\left[ {{{\lim }_{a \to {1^ - }}}\left\{ {\left( {\log (1 - a} \right)\left( {1 - {a^2}} \right) - a\left( {1 - {2^{\bf{ - 1}}}a} \right)} \right\}} \right]} = - \frac{3}{4}~?

That seems straightforward. NO?

Yeah...the question is how I actually formulate the solution.

Because what I see is https://www.wolframalpha.com/input/?i=(1/2(-infinity)*0+-3/4)
 
Initial Problem:
https://www.wolframalpha.com/input/?i=limit+(integral+from+0+to+a+of+(xln(1-x)))+as+a->1-

After taking the integral without its borders, and solving it with integration by parts:
https://www.wolframalpha.com/input/?i=limit+(ln(1-a)(a^2-1))/2+-(a(1+a/2))/2+as+a->1-

Graph of the function inside the limit:
https://www.desmos.com/calculator/mbarwveofq

I don't know how to properly solve the limit. Maybe it's no biggie, but any help is appreciated.
Assuming you can do the integral, the part which is probably giving you trouble is the (x2-1)log(1-x). Have you had L'Hôpital's rule yet? If so write that as
(x21)log(1x)=log(1x)1x21\displaystyle (x^2-1)log(1-x)\, =\, \frac{log(1-x)}{\frac{1}{x^2-1}}.

Another possibility is that you have had the limit as x goes to zero of x log(x)=0. If that is true then write the expression as
(x21)log(1x)=(x+1)ulog(u)\displaystyle (x^2-1)log(1-x)\, = -(x+1)\, u\, log(u)
where u=1-x.
 
I really don't know which is the actual question. It seems to be why:

21[lima1{(log(1a)(1a2)a(121a)}]=34 ?\displaystyle {\large{2^{ - 1}}\displaystyle\left[ {{{\lim }_{a \to {1^ - }}}\left\{ {\left( {\log (1 - a} \right)\left( {1 - {a^2}} \right) - a\left( {1 - {2^{\bf{ - 1}}}a} \right)} \right\}} \right]} = - \frac{3}{4}~?

No, that equals -1/2.



21[lima1{(log(1a)(1a2)a(1+21a)}]=34\displaystyle {\large{2^{ - 1}}\displaystyle\left[ {{{\lim }_{a \to {1^ - }}}\left\{ {\left( {\log (1 - a} \right)\left( {1 - {a^2}} \right) - a\left( {1 + {2^{\bf{ - 1}}}a} \right)} \right\}} \right]} = - \frac{3}{4}
pka said:
That seems straightforward. NO?

No, your post is not of significant help. It's condescending. It's close to: "I can do it, so you should be able to, too."


I've rewritten the part of the expression with certain grouping symbols to emphasize what the argument for the log is.


lima1 {[ln(1a)](1a2)]}\displaystyle \displaystyle\lim_{a \to {1^ - }} \ \{ [{\ln (1 - a)] ( 1 - a^2)}]\}


The above has to be shown that it equals 0. It involves an indeterminate form.


Have you had L'Hôpital's rule yet? If so, you may write that as

(1a2)ln(1a)=ln(1a)11a2\displaystyle (1 - a^2)ln(1 - a)\, =\, \dfrac{ln(1 - a)}{\frac{1}{1 - a^2}}.
 
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Yeah...the question is how I actually formulate the solution.
Because what I see is https://www.wolframalpha.com/input/?i=(1/2(-infinity)*0+-3/4)

It is well known that limx0+xlog(x)=0\displaystyle {\displaystyle\lim _{x \to {0^ + }}}x\log (x) = 0. See if you can prove that.

That is exactly what we have. See here.

You see as a1\displaystyle a\to 1^{\bf-} then (1a)0+ & (1a2)0.\displaystyle (1-a)\to 0^+~\&~(1-a^2)\to 0.
 

Yeah I eventually solved it by rewriting in such a way that by P.I.M I got -infinity/infinity, and then by using the l'Hopital once it removes the indeterminate form.
 
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