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**

Lt x^2{f(x+1)-f(x) } as x--> infinity, f(x) arc tan x.

x^2 {f(x+1)-f(x)} as x--> inf.

We know that arctan (x+1)-arctanx = arc tan { (x+1)-x]/(1+x(x+1)}= arctan (1/1+x+x^2)

Therfore Lt x^2*f{(x+1)-f(x) } = ltx^2 arctan {1/1+x+x^2} becomes infinity*0 form for x --> infinity

So we use L'Hospital's rule for arctan 1/(1+x+x^2)/ (1/x^2):

lt(x^2)' (1+x+x^2)'/(1+(1/1+x+x^2)^2 }/(-2/x^3)

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

= lt {2x^7 +..)/(x^4+...} = 2x^3 + lesser power of x

= infinity.

Therefore lt x^2 {f(x+1) - f(x) } = infinity as x--> infinity.