# limit of $\lim_{n \to \infty }\frac{n}{\ln\left ( \frac{3n}{10} \right )}$

Help me please to find: $\lim_{n \to \infty }\frac{n}{\ln\left ( \frac{3n}{10} \right )}$

Thanks.

-
It helps to think about what happens when n = 1000, n = 100000, n = 1000000000 etc – Adam Rubinson Nov 16 '12 at 16:06

$\lim_{n \to \infty }\frac{n}{\ln\left ( \frac{3n}{10} \right )}$ $=\lim_{n \to \infty }\frac{n}{\ln3+\log n-\log{10} }$

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, $$\lim_{n \to \infty }\frac{n}{\ln\left ( \frac{3n}{10} \right )} =\frac{10}3\lim_{m \to \infty }\frac{e^m}m =\frac{10}3\lim_{m \to \infty }\frac{1+\frac m{1!}+\frac {m^2}{2!}+\cdots }m$$ $$=\frac{10}3\cdot\lim_{m \to \infty }\{ \frac 1m +m+\frac m{2!}+\cdots\}=\infty$$

-
You forgot a lim. – N.U. Nov 16 '12 at 14:24
@N.U., thanks . Rectified. – lab bhattacharjee Nov 16 '12 at 14:25
@lab bhattacharjee Thank you very much – Tina Nov 17 '12 at 8:28
@Tina, welcome. Hope I could clear the idea. – lab bhattacharjee Nov 17 '12 at 8:31