The function #F,# is defined by,
# F(x)=(36^x-9^x-4^x+1)/(sqrt2-sqrt(1+cosx)), if x ne0:#
#F(x)=K, if x=0.#
Knowing that, #1+cos2theta=2cos^2theta, and, #
# 1-cos2theta=2sin^2theta," we see that, for "xne0,#
#F(x)=(36^x-9^x-4^x+1)/(sqrt2-sqrt(1+cosx)),#
#={9^x(4^x-1)-1(4^x-1)}/{sqrt2-sqrt(2cos^2(x/2))},#
#={(4^x-1)(9^x-1)}/{(sqrt2)(1-cos(x/2))},#
#={(4^x-1)(9^x-1)}/{(sqrt2)(2sin^2(x/4))},#
#=1/(2sqrt2){(4^x-1)/x}{(9^x-1)/x}(x/sin(x/4))(x/sin(x/4)),#
#=1/(2sqrt2){(4^x-1)/x}{(9^x-1)/x}((x/4*4)/sin(x/4))((x/4*4)/sin(x/4)),#
#=16/(2sqrt2){(4^x-1)/x}{(9^x-1)/x}((x/4)/sin(x/4))((x/4)/sin(x/4)),#
Now, recall that, #lim_(h to 0)(a^h-1)/h=lna, and, lim_(h to 0)sinh/h=1.#
# :. lim_(x to 0)F(x)=(4sqrt2)(ln4)(ln9)(1)(1)=(4sqrt2)(ln4)(ln9).#
#because," F is continuous at "x=0, :., lim_(x to 0)F(x)=F(0)=K,#
# rArr K=(4sqrt2)(ln4)(ln9).#
Enjoy Maths.!