Question

$\underset{x\to 0}{lim}{\left(\frac{e^{x\log (3^x-1)}-(3^x-1{)}^x\sin x}{e^{x\log x}}\right)}^{\frac{1}{x}}=$

• A. $e$
• B. $\frac{1}{e}\log 3$
• C. $e\log 3$
• D. $\log 3$

Answer 1

The correct option is B $\frac{1}{e}\log 3$

$\underset{x\to 0}{lim}{\left(\frac{e^{xlog(3^x-1)}-(3^x-1{)}^{x}sinx}{e^{xlogx}}\right)}^{\frac{1}{x}}$
$=\underset{x\to 0}{lim}{\left(\frac{(3^x-1{)}^x-(3^x-1{)}^{x}sinx}{x^x}\right)}^{\frac{1}{x}}$
$=\underset{x\to 0}{lim}{\left[{\left(\frac{3^x-1}{x}\right)}^x(1-sinx)\right]}^{\frac{1}{x}}$.
$=\underset{x\to 0}{lim}\frac{3^x-1}{x}(1-sinx{)}^{\frac{1}{x}}$.
$\underset{x\to 0}{lim}\frac{3^x-1}{x}.\underset{x\to 0}{lim}(1-sinx{)}^{\frac{1}{x}}(\because {lim}_{x\to 0}\frac{a^x-1}{x}=loga)$
$=log3\times e^{\underset{x\to 0}{lim}\frac{(1-sinx-1)}{x}}$
$=log3\times e^{-1}(\because \underset{x\to 0}{lim}\frac{sinx}{x}=1)$
$=\frac{log3}{e}$.