The derivative of arctan(ax) is a/(a^2*x^2+1)
The indefinite integeral of arctan(ax) is x*arctan(ax) - ln(a^2*x^2+1)/(2a)
The indefinite integeral, when evaluated at 0 is 0. So the limit may be expressed as:
lim {x to inf} [x*arctan(ax) - x*arctan(bx)] - [ln(a^2*x^2+1)/(2a) - ln(b^2*x^2+1)/(2b)]
The first half of the limit can be expressed as [arctan(ax) - arctan(bx)]/[1/x], to which L'Hopital's rule may be applied. Then the limit simplifies to 1/b - 1/a.
The second half of the limit simplifies into ln(a)/a - ln(b)/b + ln(x)*(1/a-1/b). Then this limit diverges to infinity.
So the combined limit also diverges to infinity.
A little more direct....
