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Division of a Polynomial by a Polynomial

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_{1}

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e^{a_{n}}

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a_{1}

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a_{n}

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Solution

The correct option is D $a_{1}$
$lim x \to \infty F (x) = L lim x \to \infty (p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + . . . . + p_{n} a_{n}^{x})^{1 / x}$
$∴ l n L = {lim}_{x \to \infty} \frac{(p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + . . . . + p_{n} a_{n}^{x})}{x}$
Using L' Hospital's rule
$l n L = lim x \to \infty \frac{p_{1} a_{1}^{x} l n a_{1} + p_{2} a_{2}^{x} l n a_{2} + . . . . + p_{n} a_{n}^{x} l n a_{n}}{p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + . . . . + p_{n} a_{n}^{x}}$ (i)
Dividing by $a_{1}^{x}$ and taking limit, we get
$lim x \to \infty {(\frac{a_{2}}{a_{1}})}^{x}, {(\frac{a_{3}}{a_{2}})}^{x},$ etc.
All vanishes as $x \to \infty$ . Therefore,
$l n = L \frac{p_{1} l n a_{1}}{p_{1}} = l n a_{1}$
or $L = a_{1}$
Hence, option 'C' is correct.

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