How do you find the limit (X-> 0) ? thank you

#(((3^X)^2+(2^X)^2)/(3^X+2^X))^(1/X)#

1 Answer
Feb 12, 2018

#sqrt(6)#

Explanation:

#a^x = exp(x*ln(a))#
#= 1 + x*ln(a) + (x*ln(a))^2/2 + (x*ln(a))^3/6 + ...#

#=> 3^x+2^x = 2 + x*(ln(3)+ln(2)) + x^2*(ln(3)^2+ln(2)^2)/2 + x^3*(ln(3)^3+ln(2)^3)/6 + ...#
#= 2 + x*ln(6) + x^2*(...#

#=> (3^x)^2 + (2^x)^2 = 3^(2x) + 2^(2x) #
#= 2 + 2*x*ln(6) + 4*x^2*(ln(2)^2+ln(3)^2)/2 + 8*x^3*(ln(3)^3+ln(2)^3)/6 + ...#

#=> (3^(2x) + 2^(2x))/(3^x+2^x) ="#
#1 + (x*ln(6)+3*x^2*...)/(2+x*ln(6)+x^2*...)#
#~~ 1+(x*ln(6))/2" (for x"->"0)"#

#"raised to the power 1/x yields : "#
#(1+(x*ln(6))/2)^((2/(x*ln(6)))*(ln(6)/2))#
#= e^(ln(6)/2)#
#= sqrt(6)#