The value of limx→∞ x[tan−1(x+1x+2)−tan−1(xx+2)]is
1/2
=limx→∞ x[tan−1(x+1x+2)−tan−1(xx+2)]is=limx→∞ x tan−1(x+1x+2−xx+21+x+1x+2xx+2)
=limx→∞ x tan−1(x+22x2+5x+4)=limx→∞⎡⎢⎣tan−1(x+22x2+5x+4)x+22x2+5x+4⎤⎥⎦×x(x+2)2x2+5x+4
Using standard limits -
=1×12=12