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

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Answer 1 · 2016-04-28T04:04:09

Gather terms into a single ratio and apply L'Hopital's rule twice to find

#lim_(x->1)(x/(x-1)-1/ln(x))=1/2#

#lim_(x->1)(x/(x-1)-1/ln(x)) = lim_(x->1)(1+1/(x-1)-1/ln(x))#

#=lim_(x->1)(1+(ln(x)-x+1)/((x-1)ln(x)))#

#=1+lim_(x->1)(ln(x)-x+1)/((x-1)ln(x))#

As the above limit is a #0/0# indeterminate form, we may apply L'Hopital's rule.

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

#=1+lim_(x->1)(1/x-1)/(1+ln(x)-1/x)#

This is another #0/0# indeterminate form, so we apply L'Hopital's rule again.

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

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

#=1+(-1)/(1+1)#

#=1/2#