1. ## limit problem_ass4_9

$lim_{x \to \infty} \frac{ln(e^{3x}-5x)}{x}$

How I can solve this ? I tried L'Hospital's but I stuck ...

2. Originally Posted by gilyos
$lim_{x \to \infty} \frac{ln(e^{3x}-5x)}{x}$

How I can solve this ? I tried L'Hospital's but I stuck ...
hmmm.... after using L.H rule once you should get

$\lim_{x \to \infty}\frac{\frac{3e^{3x}-5}{e^{3x}-5x}}{1}= \lim_{x \to \infty}\frac{3e^{3x}-5}{e^{3x}-5x}$

This is still of the form infinity over infinity.

use L.H's Rule again

$=\lim_{x \to \infty} \frac{9e^{3x}}{3e^{3x}-5}$

Finally use it one more time and use algebra to simplify.

From a different point of view note that

for large x $\ln(e^{3x}-5x) \approx \ln(e^{3x})= 3x$

so the limit is 3

3. $\lim_{x \to \infty}\frac{3e^{3x}-5}{e^{3x}-5x}= \frac{[\infty-5]}{[\infty]}$

There permission here to L'Hospital's ??

4. $\lim_{x\to\infty}\frac{3e^{3x}-5}{e^{3x}-5x}=\lim_{x\to\infty}\frac{3-\frac{5}{e^{3x}}}{1-\frac{5x}{e^{3x}}}=3$

Also,
$\lim_{x\to\infty}\frac{\log\left(e^{3x}-5x\right)}{x}=\log\left\{\lim_{x\to\infty}\left(e^ {3x}-5x\right)^\frac{1}{x}\right\}$ and $\underbrace{\left(e^{3x}-\frac{1}{2}e^{3x}\right)^{\frac{1}{x}}}_{\to e^3}\leq\left(e^{3x}-5x\right)^{\frac{1}{x}}\leq \underbrace{\left(e^{3x}+e^{3x}\right)^{\frac{1}{x }}}_{\to e^3}$

5. Originally Posted by gilyos
$lim_{x \to \infty} \frac{ln(e^{3x}-5x)}{x}$

How I can solve this ? I tried L'Hospital's but I stuck ...
For very large x, $e^{3x}$ is much larger than 5x so this has the same limit as $\frac{ln(e^{3x})}{x}= \frac{3x}{x}= 3$.

6. Originally Posted by gilyos
$lim_{x \to \infty} \frac{ln(e^{3x}-5x)}{x}$

How I can solve this ? I tried L'Hospital's but I stuck ...
Exactly what HallsOfIvy said but a little bit more formal $\ln\left(e^{3x}-5x\right)=3x+\ln\left(1-\tfrac{5x}{e^{3x}}\right)$ and since $\frac{5x}{e^{3x}}\underset{x\to\infty}\to 0$ we may conclude that $\ln\left(1-\tfrac{5x}{e^{3x}}\right)\underset{x\to\infty}{\si m}\frac{-5x}{e^{3x}}$ and the conclusion easily follows.