How do you find the limit of # [ 1/ln(x) - 1/(x-1) ] # as x approaches 1?

#lim_(xto1)[1/ln(x) - 1/(x - 1)] = ?#

Make a common denominator:

#lim_(xto1)[1/ln(x)(x - 1)/(x - 1) - 1/(x - 1)ln(x)/ln(x)] = #

#lim_(xto1)[(x - 1)/(ln(x)(x - 1)) - ln(x)/(ln(x)(x - 1))] = #

#lim_(xto1)[(x - 1 - ln(x))/(ln(x)(x - 1))]#

The above evaluates at the limit to an indeterminate form, #0/0#. Therefore, the use of L'Hôpital's rule is warranted:

Compute the first derivative of the numerator:

#(d(x - 1 - ln(x)))/dx = 1 -1/x#

Compute the first derivative of the denominator:

#(d(ln(x)(x - 1)))/dx = (x - 1)/x + ln(x)#

Make a new fraction out of the new numerator and new denominator:

#lim_(xto1)[(1 -1/x)/((x - 1)/x + ln(x))]#

Multiply by #x/x#

#lim_(xto1)[(x -1)/(x - 1 + xln(x))]#

It is still the indeterminate form #0/0#, therefore, we apply the rule, again:

Numerator:

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

Denominator:

#(d(x - 1 + xln(x)))/dx = 1 + ln(x) + x/x = 2 + ln(x)#

Here is the new expression:

#lim_(xto1)[1/(2 + ln(x))]#

The above can be evaluated at the limit:

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

Therefore, the original expression has the same limit:

#lim_(xto1)[1/ln(x) - 1/(x - 1)] = 1/2#