Question

$\underset{n\to \infty }{lim}\frac{4^{1/n}-1}{3^{1/n}-1}$ is equal to

• A. ${\log }_{4}3$
• B. $1$
• C. ${\log }_{3}4$
• D. none of these

Answer 1

The correct option is C ${\log }_{3}4$

$\underset{n\to \infty }{lim}\frac{4^{1/n}-1}{3^{1/n}-1}$
$=\underset{n\to \infty }{lim}\frac{\frac{4^{1/n}-1}{1/n}}{\frac{3^{1/n}-1}{1/n}}$
Since $n$ tends to $\infty$ $\frac{1}{n}$ tends to 0
So, applying ${lim}_{x\to 0}\frac{a^x-1}{x}={\log }_{e}a$
$=\frac{{\log }_{e}4}{{\log }_{e}3}={\log }_{3}4$