We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

#2**+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 \(-\frac{1}{1} = \boxed{-1}\).

itsyaboi Feb 4, 2019