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Let a(1)gta(2)gta(3)gt...gta(n)gt1. p(...

Let` a_(1)gta_(2)gta_(3)gt...gta_(n)gt1.`
`p_(1)gtp_(2)gtp_(3)gt...gtp_(n)gt0" such that "p_(1)+p_(2)+p_(3)+...+p_(n)=1.`
Also, `F(x)=(p_(1)a_(1)^(x)+p_(n)a_(n)^(x))^(1//x)`.
`underset(xto0^(+))limF(x)" equals "`

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Verified by Experts

The correct Answer is:
D
`Let underset(xtooo)limF(x)=L`
`:." "lnL=underset(xtooo)lim(p_(1)a_(1)^(x)lna_(1)+p_(2)a_(2)^(x)lna_(2)+...+p_(n)a_(n)^(x)lna_(n))/(p_(1)a_(1)^(x)+p_(2)a_(2)^(x)+...+p_(n)a_(n)^(x))`
Dividing by `(a_(n))^(x)` and taking `underset(xtooo)lim((a_(1))/(a_(n)))^(x),((a_(2))/(a_(n)))^(x)`,etc. vanish.
Therefore,
`lnL=(p_(n)lna_(n))/(p_(n))`
or `L=a_(n)`
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