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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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