Q.
limx→1 nxn+1-n+1xn+1ex-esinπx where n=100, is equal to
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a
5050πe
b
100πe
c
-5050πe
d
-4950πe
answer is C.
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Detailed Solution
L= limx→1 nxnx-1-xn-1ex-esinπx Put x =1+h so that as x→1, h→0. Therefore L=limh→0 h.n1+hn-1+hn-1eeh-1sinπh limh→0 n.h1+C1 nh+C2 nh2+C3 nh3+...-1+C1 nh+C2 nh2+C3 nh3+...-1πeh2eh-1hsinπhπh =-n2-C2 nπe=-2n2-nn-12πe =n2+n2πe=nn+12πe If n=100, then L=-5050πe
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