#lim_(x to 1)(1-x+ln(x))/(1-sqrt(2x-x^2)#
Both functions #f(x)=1-x+ln(x)# and #g(x)=1-sqrt(2x-x^2)# are defined and continuous on 1.
So:
Using L'Hôpital's rule :
#lim_(x to 1)(1-x+ln(x))/(1-sqrt(2x-x^2))=lim_(x to 1)((1-x+ln(x))^')/((1-sqrt(2x-x^2))^')#
#=lim_(x to 1)(-1-1/x)/(-1/2*(2-2x)*1/sqrt(2x-x^2))#
#=lim_(x to 1)((x+1)sqrt(2x-x^2))/(x-x^2)#
Note, here you can already see that the #f to oo#, but you can't know the sign. So, we need the last step to know it.
Let #X=x-1#, we have :
#lim_(X to 0) ((X+2)sqrt(2(X+1)-(X+1)^2))/(X+1-(X+1)^2)#
#=lim_(X to 0)((X+2)sqrt(2X+2-X^2-2X-1))/(X+1-X^2-2X-1)#
#=-lim_(X to 0)((X+2)sqrt(1-X^2))/(X^2+X)#
#=(-(0+2)sqrt(1-0))/0^+=(-2)/0^+#
#=-oo#
\0/ Here's our answer !
