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
limx→−∞F(x) equals
A
lnan
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B
ea1
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C
a1
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D
an
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Solution
The correct option is Dan Let limx→−∞F(x)=L ∴lnL=limx→∞p1ax1lna1+p2ax2lna2+....+pnaxnlnanp1ax1+p2ax2+....+pnaxn Dividing by (an)x and taking limx→−∞(a1an)x,(a2an)x, etc. vanish. Therefore, lnL=pnlnanpn or L=an