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

Thanks.

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It helps to think about what happens when n = 1000, n = 100000, n = 1000000000 etc –  Adam Rubinson Nov 16 '12 at 16:06
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1 Answer 1

up vote 3 down vote accepted

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

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

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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
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