Question #2849d

1 Answer
Nov 24, 2017

If the denominator is supposed to be xsqrt(x)xx, you can use L'Hopital's Rule to get an answer equal to (a^2-b^2)/(3a^2b^2)a2b23a2b2 (when a!=0a0 and b!=0b0).

Explanation:

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.