Using L'Hopital's Rule, find the a) limit as x approaches 1 of (ln(2x-1))/(x-1), b) limit as x approaches 0 from the right of (1/(e^x-1) - 1/x).

Answer 1

a) lim x-->1 (d ln(2x-1)/dx / d(x-1)/dx = lim x-->1 2/(2x-1)/1 = 2/(1)/1 = 2

b) First, reorganize the function (x - (e^x-1))/x(e^x-1) = (x-e^x+1)/(xe^x-x) --> 0/0 case so can use L'H

Take derivative of numerator and denominator

num' = 1-e^x den' = e^x+xe^x-1 still 0/0

num" = -e^x den" = e^x+e^x+xe^x = e^x(2+x)

So, lim -->0 -e^x/e^x(2+x) = -1/(1(2+0)) = -1/2