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Question

Let a1>a2>a3>....>an>1.
p1>p2>p3>....>pn>0 such that p1+p2+p3+...+pn=1.
Also, F(x)=(p1ax1+p2ax2+...+pnaxn)1/x
limxF(x) equals

A
lna1
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B
ean
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C
a1
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D
an
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Solution

The correct option is D a1
limxF(x)=Llimx(p1ax1+p2ax2+....+pnaxn)1/x
lnL=limx(p1ax1+p2ax2+....+pnaxn)x
Using L' Hospital's rule
lnL=limxp1ax1lna1+p2ax2lna2+....+pnaxnlnanp1ax1+p2ax2+....+pnaxn (i)
Dividing by ax1 and taking limit, we get
limx(a2a1)x,(a3a2)x, etc.
All vanishes as x. Therefore,
ln=Lp1lna1p1=lna1
or L=a1
Hence, option 'C' is correct.

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