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

1 Answer

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)#

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