Question

How do you evaluate the limit `lim (3^x-2^x)/x` as `x->0`?

Answer 1

`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)`