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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)`.
`lim_(xtooo) F(x)" equals "`

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The correct Answer is:
C

`underset(xtooo)limF(x)=L=underset(xtooo)lim(p_(1)a_(1)^(x)+p_(2)a_(2)^(x)+...+p_(n)a_(n)^(x))^(1//x)" "(oo^(0)" form")`
`:." "lnL=underset(xtooo)lim(log_(e)(p_(1)a_(1)^(x)+p_(2)a_(2)^(x)+...+p_(n)a_(n)^(x)))/(x)`
Using L'Hospital's rule
`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))" "(1)`
Dividing by `a_(1)^(x)` and taking limit, we get
`(becauseunderset(xtooo)lim(a_(2)/(a_(1)))^(x)=0,underset(xtooo)lim((a_(3))/(a_(1)))^(x)=0" etx.")`
`lnl=(p_(1)lna_(1))/(p_(1))=lna_(1)`
or `L=a_(1)`
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