The value of limx→∞ 2x1/2+3x1/3+4x1/4+…nx1/n(2x−3)1/2+(2x−3)1/3+…+(2x−3)1/n is
0
2
13
Let
l=limx→∞ 2x1/2+3x1/3+4x1/4+…+nx1/n(2x−3)1/2+(2x−3)1/3+…+(2x−3)1/n
Then,
l=limh→0 2h−1/2+3h−1/3+4h−1/4+…+nh−1/n(2−3h)1/2h−1/2+(2−3h)1/3h−1/3+…+(2−3h)1/nh−1/n, where h=1x.
⇒ l=limh→0 2+3h12−13+4h12−14+…+nh1 12 n(2−3h)2+(2−3h)13h12−13+…+(2−3h)nh12−1n
⇒ l=22=2