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**

To find the Limit x^2{f(x+1)-f(x) } as x-> infinity.

Consider the function, x^2{arctg(x+1)-arctg(x}

x^2arctg{[(x+1)-x]/(1+x(x+1))}

x^2arctg{1/x^2+x+1}.

As x->infinity, x^2 ->infinite and the other arctg function tends to zero becoming a **0*infinity** type, which is an indefinite form.Therefore,we use L'Hospital's rule for the case:

We shall convert the function into a **0/0** form

Convert the numerator x^2 to ** 1/(1/x^2)**.

x^2arctg{1/x^+x+1} = arctg{1/x^+x+1}/{**1/(1/x^2)**}

The numerator's differential coefficient is:

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

The differential coefficient of the denominator is

(1/x^2)'= -2/x^3....(2).

Eq(1) / Eq(2) gives after reduced by -1 in both numerator and denominator:

x^3(2x+1)/{(x^2+x+1)^2+1} =

={2+1/x}/{2{1+2/x+1/x^2+..} = (2+0)/{2(1+0 terms}=** 1** as x goes infinite.

Therefore,

Limit x-> infinity [arctg(x+1)-arctg(x)] = 1 or

Limitx->infinity (f(x+1)-f(x)) = 1, where f(x)= arctg(x)