$\displaystyle lim_{x \to \infty} \frac{ln(e^{3x}-5x)}{x} $
How I can solve this ? I tried L'Hospital's but I stuck ...
hmmm.... after using L.H rule once you should get
$\displaystyle \lim_{x \to \infty}\frac{\frac{3e^{3x}-5}{e^{3x}-5x}}{1}= \lim_{x \to \infty}\frac{3e^{3x}-5}{e^{3x}-5x}$
This is still of the form infinity over infinity.
use L.H's Rule again
$\displaystyle =\lim_{x \to \infty} \frac{9e^{3x}}{3e^{3x}-5}$
Finally use it one more time and use algebra to simplify.
From a different point of view note that
for large x $\displaystyle \ln(e^{3x}-5x) \approx \ln(e^{3x})= 3x$
so the limit is 3
$\displaystyle \lim_{x\to\infty}\frac{3e^{3x}-5}{e^{3x}-5x}=\lim_{x\to\infty}\frac{3-\frac{5}{e^{3x}}}{1-\frac{5x}{e^{3x}}}=3$
Also,
$\displaystyle \lim_{x\to\infty}\frac{\log\left(e^{3x}-5x\right)}{x}=\log\left\{\lim_{x\to\infty}\left(e^ {3x}-5x\right)^\frac{1}{x}\right\}$ and $\displaystyle \underbrace{\left(e^{3x}-\frac{1}{2}e^{3x}\right)^{\frac{1}{x}}}_{\to e^3}\leq\left(e^{3x}-5x\right)^{\frac{1}{x}}\leq \underbrace{\left(e^{3x}+e^{3x}\right)^{\frac{1}{x }}}_{\to e^3}$
Exactly what HallsOfIvy said but a little bit more formal $\displaystyle \ln\left(e^{3x}-5x\right)=3x+\ln\left(1-\tfrac{5x}{e^{3x}}\right)$ and since $\displaystyle \frac{5x}{e^{3x}}\underset{x\to\infty}\to 0$ we may conclude that $\displaystyle \ln\left(1-\tfrac{5x}{e^{3x}}\right)\underset{x\to\infty}{\si m}\frac{-5x}{e^{3x}}$ and the conclusion easily follows.