The limit #lim_{x->0}(a arctan(sqrt(x)/a)-b arctan(sqrt(x)/b))/(xsqrt(x))# is a #0/0# indeterminate form, so L'Hopital's Rule can be used.
The derivative of the numerator is
#a/(1+x/a^2) * 1/(2a sqrt(x))-b/(1+x/b^2) * 1/(2b sqrt(x))#
#=a^2/(2a^2sqrt(x)+2xsqrt(x))-b^2/(2b^2sqrt(x)+2xsqrt(x))#
#=a^2/(2sqrt(x)(a^2+x))-b^2/(2sqrt(x)(b^2+x))#
#=(a^2*(b^2+x)-b^2*(a^2+x))/(2sqrt(x)(a^2+x)(b^2+x))#
#=((a^2-b^2)x)/(2sqrt(x)(a^2+x)(b^2+x))#
#=((a^2-b^2)sqrt(x))/(2(a^2+x)(b^2+x))#.
The derivative of the denominator is #3/2 sqrt(x)#
Therefore,
#lim_{x->0}(a arctan(sqrt(x)/a)-b arctan(sqrt(x)/b))/(xsqrt(x))#
#=lim_{x->0}(((a^2-b^2)sqrt(x))/(2(a^2+x)(b^2+x)))/(3/2 sqrt(x))#
#=lim_{x->0}((a^2-b^2))/(3(a^2+x)(b^2+x))=(a^2-b^2)/(3a^2b^2)# when #a!=0# and #b!=0#.
