Evaluate the limit of x^2[f(x+1)-f(x)] x-->infinite, if f(x)=arctanx

We'll substitute f(x+1) and f(x) in the expression
above:

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

We'll substitute x by
infinite and we'll get the indetermination case,
"0/0".

We'll apply L'Hospital's rule to calculate the
limit:

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 (f'/g')= 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 factorize both 
numerator and denominator by "x^4".

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

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