Calculate lim x^2(f(x+1)-f(x)), x->infinity, f(x)=arctg x.

lim x^2*[atctg(x+1)-arctg x]=

=lim [atctg(x+1)-arctg x]/(1/x^2)

If we'll try to calculate this limit we'll see that we have to deal with an indetermination case, "0/0".

For this reason, we've chosen to calculate the limit, using L'Hospital's rule, which says:

lim (f/g)= lim (f'/g')

If f(x)=arctg(x+1)-arctg x, then

f'(x)={1/[1+(x+1)^2]}-1/(1+x^2)

If g(x)=1/x^2, then g'(x)=-2/x^3

lim (f'/g')= lim {1/[1+(x+1)^2] -1/(1+x^2)}/(-2/x^3)=

=lim (1+x^2-1-x^2-2x-1)/[(1+x^2)*1+(x+1)^2]*limx^3/-2

lim (-2x^4-x^3)/-2(1+x^2)(x^2+2x+2)

We'll draw out the common factor "x^4", at numerator and denominator, same time.

lim x^4(-2-1/x)/-2x^4(1+1/x^2)(1/x^2+2/x^3+2/x^4)

lim 1/x=0, x->infinity

lim 1/x^2=0,

lim 2/x^3=0 and lim 2/x^4=0

lim (-2-1/x)/-2(1+1/x^2)(1/x^2+2/x^3+2/x^4)=-2/-2=1