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Can someone give me please some guidance hoe to solve the following limit, without using L'Hopital rule?

$$\lim\limits_{n \to \infty } \frac{n}{\ln\left(\frac{3n}{5}\right)}$$

Thanks a lot!

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    $\begingroup$ Hint : $ln(\frac{3n}{5})=ln(3)-ln(5)+ln(n)$ $\endgroup$
    – Peter
    Nov 25 '14 at 19:58
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Since $n>\frac{3n}{5}$ for all $n>0$ and by the monotonicity of $\ln(x)$, we have $\ln(n)>\ln\left(\frac{3n}{5}\right)$. So $$ \frac{n}{\ln(n)}<\frac{n}{\ln\left(\frac{3n}{5}\right)}. $$ So what do you know about the lower bound?

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  • $\begingroup$ Thanks! But we didn't learn integrals yet... $\endgroup$
    – bimba bi
    Nov 25 '14 at 20:09
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Just use some algebra and the rules of division and multiplication inside logarithms:

$lim_{n\rightarrow\infty}\frac{n}{\ln(\frac{3n}{5})}=lim_{n\rightarrow\infty}\frac{n}{\ln(3/5)+\ln(n)}=lim_{n\rightarrow\infty}(\frac{\ln(n)}{n}+\frac{3/5}{n})^{-1}=({0^++0^+})^{-1}=+\infty$

This might feel a little sketchy as a proof, but each step can be well supported and should be sufficient for your present needs.

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For variety, here's another way to approach it:

$$\lim_{n \to \infty } \frac{n}{\ln(\frac{3n}{5})} = \lim_{n \to \infty } \frac{5}{3}\frac{\frac{3n}{5}}{\ln(\frac{3n}{5})} = \frac{5}{3}\lim_{n \to \infty } \frac{\frac{3n}{5}}{\ln(\frac{3n}{5})}$$

Now substitute $k = \frac{3n}{5}$, so that the limit becomes

$$\frac{5}{3} \lim_{k \to \infty } \frac{k}{\ln k} \ .$$

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the $3$ and $5$ are not important that you will see soon. what is important is even though $\ln(x)$ is large for $x$ large but the quotient $\frac{x}{\ln(x)}$ is very large, in other words $x$ goes to $\infty$ much faster than $\ln(x).$ you can verify this by putting large values for $x,$ for example, $\frac{100}{\ln(100)} = 21.7, \frac{\ln(1000)}{1000} = 144.8, \cdots$ you see the pattern.

now back to your question: $\frac{n}{\ln(3n/5)} = \frac{n}{\ln(n) + \ln(3)-\ln(5)} = \frac{n}{\ln(n)} + \cdots$ which is very large for $x$ large. another way saying this is the $\lim_{n \to \infty}\frac{n}{\ln(3n/5)}$ does not exist.

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