without using l'hopital rule

Question

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!

Answer 1

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?

Answer 2

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.

Answer 3

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.