limit x tends to 1 (1-x2)^1/log(1-x)

Question

answer -e

Answer 1

I get   e   not  -e.

 

My method let  z = 1-x   The desired limit is then z → 0⁺   of  

 

   (2-z)^1/lnz  z^1/lnz

 

  The second term can be written as exp[ln( z^1/ln(z) ) ]  = exp[ ( 1/lnz) lnz ) ] = exp [1] =  e

 

In the limit, the first term becomes ~ 2⁰  which is equal to 1  

 

Thus overall the limit is   equal to e.