How do you evaluate #( ( x / (x-1 ) - (1 / lnx ) )# as x approaches 1+?

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Answer 1 · 2017-11-16T16:36:09

#1/2#

#x/(x-1)-1/(lnx)=(xlnx-(x-1))/((x-1)lnx)=(xlnx-x+1)/(xlnx-lnx)#

Plugging in 1 gives the indeterminate form #0/0#

Using L'Hospital's Rule:

Differentiate numerator and denominator.

#d/dx(xlnx-x+1)=lnx#

#d/dx(xlnx-lnx)=lnx+1/x*x-1/x=lnx+1-1/x#

We still have the form #0/0#, so differentiate again.

#d/dx(lnx)=1/x#

#d/dx(lnx+1-1/x)=1/x+1/x^2#

#:.#

#(1/x)/(1/x+1/x^2)=(x/x)/(x/x+x/x^2)=1/(1+1/x)#

Plugging in 1:

#1/(1+1/1)=1/(1+1)=1/2#

#:.#

#lim_(x->1^(+))((xlnx-x+1)/(xlnx-lnx))=1/2#