Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 51

lim_(x->0^+) (1/x-1/(e^x-1))
Plug-in x=0 to the function.
lim_(x->0^+) (1/x-1/(e^x-1)) = 1/0-1/(e^0-1)=oo - oo
Since the result is indeterminate, to take its limit apply L'Hospital's Rule. To do so, expression the function as one fraction.
lim_(x->0^+) (1/x-1/(e^x-1)) = lim_(x->0^+) ((e^x-1)/(x(e^x-1)) - x/(x(e^x-1)))=lim_(x->0^+) (e^x-x-1)/(x(e^x-1))
Then, take the derivative of the numerator and denominator.
lim_(x->0^+) ((e^x-x-1)')/((x(e^x-1))')=lim_(x->0^+) (e^x-1-0)/(x*e^x + 1*(e^x-1)) = lim_(x->0^+) (e^x-1)/(xe^x + e^x-1)
And, plug-in x=0.
lim_(x->0^+)(e^x-1)/(xe^x+e^x-1)= (e^0-1)/(0*e^0+e^0-1) = 0/0
Since the result it still indeterminate, apply L'Hospitals Rule again. Take the derivative of the numerator and denominator.
lim_(x->0^+) ((e^x-1)')/((xe^x+e^x-1)') = lim_(x->0^+) (e^x-0)/(x*e^x+1*e^x+e^x-0) = lim_(x->0^+) e^x/(xe^x + 2e^x) = lim_(x->0^+)1/(x+2)
And, plug-in x=0.
lim_(x->0^+)(1/x+2)=1/(0+2)=1/2

Therefore, lim_(x->0^+) (1/x -1/(e^x-1)) = 1/2 .