How do you evaluate the limit #lim (3^x-2^x)/x# as #x->0#? Calculus Limits Infinite Limits and Vertical Asymptotes 1 Answer Cesareo R. May 10, 2017 #log_e(3/2)# Explanation: #(3^x-2^x)/x=2^x((3/2)^x-1)/x = 2^x(((3/2)^(0+x)-(3/2)^0)/x)# so #lim_(x->0)(3^x-2^x)/x=lim_(x->0)2^x lim_(x->0)(((3/2)^(0+x)-(3/2)^0)/x)=1 d/(dx)(3/2)^x]_(x=0) = log_e(3/2)# Answer link Related questions How do you show that a function has a vertical asymptote? What kind of functions have vertical asymptotes? How do you find a vertical asymptote for y = sec(x)? How do you find a vertical asymptote for y = cot(x)? How do you find a vertical asymptote for y = csc(x)? How do you find a vertical asymptote for f(x) = tan(x)? How do you find a vertical asymptote for a rational function? How do you find a vertical asymptote for f(x) = ln(x)? What is a Vertical Asymptote? How do you find the vertical asymptote of a logarithmic function? See all questions in Infinite Limits and Vertical Asymptotes Impact of this question 17077 views around the world You can reuse this answer Creative Commons License