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Expansions to Remove Indeterminate Form

limx → 0 [2x+...

Question

$lim x \to 0 {[\frac{2^{x} + 2^{2 x} + 2^{3 x}}{3}]}^{\frac{1}{x}}$ is equal to

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Solution

This is $1^{\infty} form$
$= e^{lim x \to 0 \frac{1}{x} [\frac{2^{x} + 2^{2 x} + 2^{3 x}}{3} - 1]}$
$= e^{lim x \to 0 \frac{1}{x} [\frac{2^{x} + 2^{2 x} + 2^{3 x} - 3}{3}]}$
$= e^{lim x \to 0 ([\frac{2^{x} - 1}{x}] + 2 [\frac{2^{2 x} - 1}{2 x}] + 3 [\frac{2^{3 x} - 1}{3 x}]) . \frac{1}{3}}$
$= e^{lim x \to 0 [\frac{ln 2 + 2 ln 2 + 3 ln 2}{3}]}$
$= e^{lim x \to 0 [\frac{2 ln 2}{1}]}$
$= e^{(2 ln 2)}$
$= e^{(ln 4)} = 4$

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then the ordered pair

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Expansions to Remove Indeterminate Form

Standard XII Mathematics

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