Question

The value of $ \displaystyle \lim_{x \to 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-cos\,2x} $ is

• A. $ \frac{\left(log\,3\right)\,\left(log\,5\right)}{2} $
• B. $ 2 \,(log \,3) \,(log \,5) $
• C. $\frac{log\,3 + log\,5}{2}$
• D. None of these

Answer 1

Answer: (A)

$\displaystyle \lim _{x \rightarrow 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-\cos 2 x}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{\left(3^{x}-1\right)\left(5^{x}-1\right)}{1-1+2 \sin ^{2} x}$
$= \displaystyle\lim _{x \rightarrow 0}\left(\frac{3^{x}-1}{x}\right)\left(\frac{5^{x}-1}{x}\right)\left(\frac{x^{2}}{2 \sin ^{2} x}\right)$
$=\frac{1}{2}(\log 3)(\log 5) \cdot 1$
$\left[\because \displaystyle\lim _{x \rightarrow 0}\left(\frac{a^{x}-b^{x}}{x}\right)=\log \left(\frac{a}{b}\right)\right]$