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Inverse of a Function

Let a1 > a2...

Question

Let $a_{1} > a_{2} > a_{3} > . . . . > a_{n} > 1.$
$p_{1} > p_{2} > p_{3} > . . . . > p_{n} > 0$ such that $p_{1} + p_{2} + p_{3} + . . . + p_{n} = 1.$
Also, $F (x) = (p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + . . . + p_{n} a_{n}^{x})^{1 / x}$
$lim x \to - \infty F (x)$ equals

ln a_{n}

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e^{a_{1}}

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a_{1}

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a_{n}

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Solution

The correct option is D $a_{n}$
Let ${lim}_{x \to - \infty} F (x) = L$
$∴ l n L = {lim}_{x \to \infty} \frac{p_{1} a_{1}^{x} l n a_{1} + p_{2} a_{2}^{x} l n a_{2} + . . . . + p_{n} a_{n}^{x} l n a_{n}}{p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + . . . . + p_{n} a_{n}^{x}}$
Dividing by $(a_{n})^{x}$ and taking $lim x \to - \infty {(\frac{a_{1}}{a_{n}})}^{x}, {(\frac{a_{2}}{a_{n}})}^{x}$ , etc. vanish.
Therefore,
$l n L = \frac{p_{n} l n a_{n}}{p_{n}}$
or $L = a_{n}$

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