Limit $\lim_{x \to \infty} \ln x - x$ Consider $f(x) = \ln \ x  - x$ where $\ln \ x$ refers to base $e$ logarithm. Find limit of $f(x)$ when $x$ goes to infinity.
My try: I know that $f'(x) = \frac{1}{x} - 1$ but what's the next step?
Please Help!
 A: Let $f(x)=e^{\ln x-x}$, hence $f(x)=\frac{x}{e^x} \to 0$  for $x \to \infty$.
Thus $ \ln x-x \to - \infty$ for $x \to \infty$.
A: $\ln(x)$ is a function that is strictly increasing on $\mathbb R^+$, but it "flattens out". 
$x$ is also strictly increasing, but it keeps a constant positive slope.
This means that $x$ will increase infinitely more than $\ln(x)$.
Thus, $\ln(x)-x \to -\infty$ as $x\to\infty$.
Alternative method using the derivative you found
You know that $f'(x) = \frac1x - 1$, and what we can see from this is that when $x\to\infty$, then $\frac1x \to 0$, and as we subtract 1, the slope will always be negative, and thus the function will tend towards $-\infty$.
A: You can use your observation for the derivative. Note that 
$$ f'(x) = \frac 1x - 1 \to -1, \quad x\to \infty $$
Hence, we can choose an $N$ such that 
$$ f'(x) < -\frac 12, \qquad x \ge N. $$
This allows us to estimate
$$ f(x) = f(N) + \int_N^x f(y)\, dy < f(N) - \frac 12(x-N) \to -\infty, \qquad x \to \infty $$
A: We have $\ln(x) - x$. Factoring out $x$, we get:
$x (\frac{\ln(x)}{x} - 1)$
We know as $x \to \infty$, the first term becomes $\infty$
What about $(\frac{\ln(x)}{x} - 1)?$
If we consider $(\frac{\ln(x)}{x})$, we get an indeterminate form $\frac{\infty}{\infty}$. Applying L'Hopital, we can rewrite $(\frac{\ln(x)}{x})$ in terms of derivative as $(\frac{\frac{1}{x}}{1})$. Evaluating this derivative as $x \to \infty$ gives us $0$.
So we have:
$\infty(0  - 1) = \infty(-1) = -\infty$
