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# lim ( (x*lnx)/(1-x) , x=1)

Feb 4, 2019

#1
+99309
+1

This is not the algebraic answer that I am sure you need, but....

Feb 4, 2019
#2
+79
+1

First, multiply by the conjugate of 1-x, which is $$\frac{1-x^2}{1+x}$$

Then, simplifying things, we get $$\lim\limits_{x\rightarrow1}(-\frac{x \ln x}{x-1})$$.

Which then turns into this: $$-\lim _{x\to \:1}\left(\frac{\frac{1}{-1+x}\ln \left(x\right)}{\frac{1}{x}}\right) \\ \frac{\lim _{x\to \:1}\left(\frac{1}{-1+x}\ln \left(x\right)\right)}{\lim _{x\to \:1}\left(\frac{1}{x}\right)}$$

Using L'Hopital's rule and simplifying, we get that this equation is equal to 1.

We also need to find the value of $$\lim _{x\to \:1}\left(\frac{1}{x}\right)$$ which is easily equal to 1.

Therefore your answer is $$-\frac{1}{1} = \boxed{-1}$$.

Feb 4, 2019
#3
+99309
+2

I always forget about L'hopital's rule - thanks for you answer and thanks for reminding me.

Bur....

$$\displaystyle\lim_{x\rightarrow 1}\;\;\frac{xlnx}{1-x}\\ =\displaystyle\lim_{x\rightarrow 1}\;\;\frac{lnx+1}{-1}\\ =-1$$

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Feb 5, 2019
edited by Melody  Feb 5, 2019
#4
+79
+1

whoops, i just realized :P

itsyaboi  Feb 5, 2019