Question

Evaluate:
$\rm \displaystyle \lim_{x \to 0}\ \frac{a^{3x}-a^{2x}-a^{x}+1}{x\sin x}$

Answer 1

We know that

$\lim_{x \to 0} \bigg(\frac{ {a}^{x} - 1}{x}\bigg) = log \: a \\ \\ \lim_{x \to 0} \bigg(\frac{sin \: x}{x}\bigg) =1$

${a^{3x}-a^{2x}-a^{x}+1} = {a}^{2x} ( {a}^{x} - 1) - 1( {a}^{x} - 1) \\ \\ = ( {a}^{x} - 1)( {a}^{2x} - 1)$
$\lim_{x \to 0}\ \frac{a^{3x}-a^{2x}-a^{x}+1}{x\sin x} \\ \\ = \lim_{x \to 0} \frac{ ( {a}^{x} - 1)( {a}^{2x} - 1)}{x \: sin \: x} \\ \\ = 2\lim_{x \to 0} \frac{ ( {a}^{x} - 1)( {a}^{2x} - 1)}{2 {x}^{2} \frac{sin \: x}{x} } \\ \\ = \frac{ 2\lim_{x \to 0} (\frac{ {a}^{x} - 1}{x}) .\lim_{x \to 0} (\frac{ {a}^{2x} - 1}{2x})}{\lim_{x \to 0} \frac{sin \: x}{x} } \\ \\ = 2 \bigg(\frac{log \: a \times log \: a}{1} \bigg) \\ \\ \lim_{x \to 0}\ \frac{a^{3x}-a^{2x}-a^{x}+1}{x\sin x}= 2 ( {log \: a})^{2}$
Hope it helps you.