without using l'hopital rule 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!
 A: 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?
A: 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.
A: 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
$$\frac{5}{3} \lim_{k \to \infty } \frac{k}{\ln k} \ .$$
A: 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.
