Question

Evaluate:
$\rm \displaystyle \lim_{x \to 0}\ \frac{15^{x}-5^{x}-3^{x}+1}{x\sin x}$

Answer 1

We know that

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

${15^{x}-5^{x}-3^{x}+1} = {3}^{x}{5^{x}-5^{x}-3^{x}+1} \\ \\= {5^{x}({3}^{x}-1)-1(3^{x}-1)}= ( {5}^{x} - 1)( {3}^{x} - 1)$
$\lim_{x \to 0}\ \frac{15^{x}-5^{x}-3^{x}+1}{x\sin x}\\ \\ = \lim_{x \to 0} \frac{ ( {5}^{x} - 1)( {3}^{x} - 1)}{x \: sin \: x} \\ \\ = \lim_{x \to 0} \frac{ ( {5}^{x} - 1)( {3}^{x} - 1)}{ {x}^{2} \frac{sin \: x}{x} } \\ \\ = \frac{ \lim_{x \to 0} (\frac{ {5}^{x} - 1}{x}) .\lim_{x \to 0} (\frac{ {3}^{x} - 1}{x})}{\lim_{x \to 0} \frac{sin \: x}{x} } \\ \\ = \bigg(\frac{log \: 5 \times log \: 3}{1} \bigg) \\ \\ \lim_{x \to 0}\ \frac{15^{x}-5^{x}-3^{x}+1}{x\sin x}= ( {log \: 5.log\:3})$
Hope it helps you.