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Question

Let a1>a2>a3>>an>1; p1>p2>p3>>pn>0 be such that p1+p2+p3++pn=1. If F(x)=(p1ax1+p2ax2++pnaxn)1/x, then

A
limx0F(x)=ap11ap22apnn
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B
limx0F(x)=ap11+ap22++apnn
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C
limxF(x)=a1
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D
limxF(x)=an
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Solution

The correct option is D limxF(x)=an
limx0F(x)=limx0(p1ax1+p2ax2++pnaxn)1/x
=elimx0p1ax1+p2ax2++pnaxn1x
=elimx0(p1ax1lna1+p2ax2lna2++pnaxnlnan)
=e(p1lna1+p2lna2++pnlnan)
=e(lnap11+lnap22++lnapnn)
=e(lnap11ap22apnn)
=ap11ap22apnn

L=limxF(x)=limx(p1ax1+p2ax2++pnaxn)1/x (0 form)
lnL=limxln(p1ax1+p2ax2++pnaxn)x
Using L'Hopitals' rule, we get
lnL=limxp1ax1lna1+p2ax2lna2++pnaxnlnanp1ax1+p2ax2++pnaxn

Multiplying and dividing by ax1 in RHS, we get
(a2a1)x,(a3a1)x, etc. all vanish as x because a1>a2>a3>an>1
lnL=p1lna1p1=lna1
Hence, lnL=lna1
L=a1


Let limxF(x)=L1
lnL1=limxp1ax1lna1+p2ax2lna2++pnaxnlnanp1ax1+p2ax2++pnaxn
Multiplying and dividing by (an)x in RHS (a1an)x,(a2an)x, etc. vanish as x
lnL1=pnlnanpn
L1=an

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