Find the $\lim_{x\to \infty} (\ln(x)-x)$ $\lim_{x\to \infty}(\ln(x)-x)$
I would like to pull the x out. However, I am unsure if my algebra is flawed or their is some certain identity I am oblivious of. 
The limit looks like this when the x is factored out. But, how was that done? 
$\lim_{x\to \infty} x\left(\frac{\ln(x)}{x}-1\right)$
How was this done? 
Thanks in advance. 
 A: Hint:
$$\ln(x)-x=\ln\left(\frac{x}{e^x}\right)$$
And $$\frac{x}{e^x}\to0\quad\text{as}\quad x\to\infty$$
A: Since $x=\ln(e^x)$, we get that this limit is $$\lim_{x\to\infty}(\ln(x)-\ln(e^x))=-\lim_{x\to\infty}\ln\left(\frac{e^x}{x}\right).$$ Can you solve this one?
A: Here's what was done. 
First, the top and the bottom were each multiplied by $x$, which is the same as multiplying by $\frac{x}{x}=1$:  
$$\lim_{x\to \infty}\frac{(\ln(x)-x)}{1}=\lim_{x\to \infty}\frac{x(\ln(x)-x)}{x}=\lim_{x\to \infty}{x(\frac{\ln(x)}{x}-\frac{x}{x})}\lim_{x\to \infty}{x(\frac{\ln(x)}{x}-1)}$$
A: $$e^{\ln x-x}=\frac x{e^x}$$.
Let $f(x)=x/e^x$. Be because $\ln x$ is continuous
$$\lim_{x\to\infty}f(x)=0$$
implies
$$\lim_{x\to\infty}\ln x-x=\lim_{x\to\infty}\ln f(x)=-\infty.$$
A: I presume that the OP is interested in evaluation of this limit, and not specifically on the legitimacy of writing $\log x-x=\log x\left(\frac{\log x}{x}-1\right)$, which is straightforward use of the distributive law.
Inasmuch as other users have posted a host of ways forward that are viable, I though it would be instructive to see yet another way to evaluate this limit.  To that end, we use herein the definition of the logarithm function
$$\log x=\int_1^x \frac{1}{t}\,dt$$
Then, we observe that for $x>1$
$$\begin{align}
\log x-x&=\int_1^x \frac{1-t}{t}\,dt-1\\\\
&\le (x-1)\left(\frac{1-x}{x}\right)-1\\\\
&=1-x-\frac1x\\\\
&\to -\infty\,\,\text{as}\,\,x\to \infty
\end{align}$$
and we are done!
A: For every $M < 0$, if $x > \max \{ 16, M^{2}/16 \}$ then
$$
\log x - x \leq \int_{t=1}^{x}t^{-1/2} - x = 2x^{1/2} - 2 - x < x^{1/2}(2 - x^{1/2}) < -2x^{1/2} < M;
$$
so the map $x \mapsto \log x - x$ diverges to negative infinity as $x$ grows indefinitely.
