Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 49

lim_(x->1) (x/(x-1) - 1/(ln(x)))
To solve, plug-in x=1.
lim_(x->1) (x/(x-1) - 1/(ln(x))) = 1/(1-1)-1/ln(1)=1/0-1/0=oo-oo
Since the result is indeterminate, we can apply the L'Hospital's Rule. To do so, subtract the two rational expressions.
lim_(x->1) (x/(x-1)-1/(ln(x))) = lim_(x->1) ((xln(x))/((x-1)ln(x)) - (x-1)/((x-1)ln(x))) = lim_(x->1) (xln(x) - x + 1)/((x-1)ln(x))
Now that we have one rational function, take the derivative of the numerator and denominator.
lim_(x->1) ((xln(x) - x + 1)')/(((x-1)ln(x))')= lim_(x->1) (x*1/x + 1*ln(x) - 1 + 0)/((x - 1)*1/x + 1*ln(x))= lim_(x->1) (1+lnx-1) /(1-1/x+ln(x))=lim_(x->1) ln(x)/(1-1/x+ln(x))
Then, plug-in x = 1.
= ln(1)/(1-1/1+ln1) = 0/0
Since the result is still indeterminate, apply the L'Hospital's Rule again. So, take the derivative of the numerator and denominator.
lim_(x->1) ((ln(x))')/((1-1/x+ln(x))') = lim_(x->1) (1/x)/(0+1/x^2+1/x)= lim_(x->1) (1/x)/(1/x^2+1/x)= lim_(x->1) x/(1+x)
Then, plug-in x=1.
=1/(1+1) = 1/2

Therefore, lim_(x->1) (x/(x-1)-1/ln(x)) = 1/2 .