Help me please to find: $\lim_{n \to \infty }\frac{n}{\ln\left ( \frac{3n}{10} \right )}$
Thanks.
Help me please to find: $\lim_{n \to \infty }\frac{n}{\ln\left ( \frac{3n}{10} \right )}$
Thanks.
$\lim_{n \to \infty }\frac{n}{\ln\left ( \frac{3n}{10} \right )}$ $=\lim_{n \to \infty }\frac{n}{\ln3+\log n-\log{10} }$
This is of the form $\frac {\infty}{\infty}$
So, we can apply L'Hospital's Rule,
$\lim_{n \to \infty }\frac{n}{\ln\left ( \frac{3n}{10} \right )} =\lim_{n \to \infty }\frac{n}{\ln3+\log n-\log{10} } =\lim_{n \to \infty }\frac1{\frac 1n}=\lim_{n \to \infty } n=\infty$
Alternatively without using L'Hospital's Rule,
let $\ln\left ( \frac{3n}{10} \right )=m,n=\frac {10}3 e^m$ and $m\to \infty$ as $n\to \infty$
So, $\lim_{n \to \infty }\frac{n}{\ln\left ( \frac{3n}{10} \right )} =\frac{10}3\lim_{m \to \infty }\frac{e^m}m =\frac{10}3\lim_{m \to \infty }\frac{1+\frac m{1!}+\frac {m^2}{2!}+\cdots }m$ $=\frac{10}3\cdot\lim_{m \to \infty }\{ \frac 1m +m+\frac m{2!}+\cdots\}=\infty $