Limit of $x - \ln(x)$ as $x$ approaches $+\infty$ To evaluate the limit of an even larger expression
$$
\lim_{x \to +\infty} \frac{\ln(\ln x)}{\ln(x - \ln x)}
$$
I need to evaluate part of the denominator to determine whether I could apply L'Hôpital's Rule
$$
\lim_{x \to +\infty} x - \ln(x)
$$
The problem is that I can't seem to manipulate the expression to the indeterminate forms $0/0$ or $\infty/\infty$. I was thinking of multiplying by $x/x$
$$
\lim_{x \to +\infty} \frac{x^2 - x\ln(x)}{x}
$$
But then I get another indeterminate form $\infty - \infty$ in the numerator. I was also thinking that $x$ grows much faster than $\ln x$, so the limit obviously tends to $+\infty$ but I don't think that would fly with most people :)
 A: If you may use derivatives, you can define
$$f(x) := x - \ln(x) $$
and note that $f(1) = 1$, hence $x>\ln(x)$ when $x = 1$.
Now compute its derivative
$$f'(x) = 1 - \frac1x. $$
It should be clear that $f'(x)>0$ for all $x>1$ so the function is growing, i.e. the difference between $x$ and $\ln x$ gets larger and since $x> \ln x$ when $x = 1$ we can conclude that the difference increases without bound.
A: With asymptotic analysis, it's very simple:
$x-\ln x\sim_\infty x$, hence $\ln(x-\ln x)\sim_\infty\ln x$, so that $\ln(\ln x)=o(\ln x)=o(\ln(x-\ln x)$ and finally
$$\frac{\ln\ln x}{\ln(x-\ln x)}\to 0 \enspace\text{as}\enspace x\to +\infty$$
A: The problem is very easy once we accept the standard limit $$\lim_{x \to \infty}\frac{\log x}{x} = 0$$ Clearly we have
\begin{align}
L &= \lim_{x \to \infty}\frac{\log \log x}{\log(x - \log x)}\notag\\
&= \lim_{x \to \infty}\dfrac{\log \log x}{\log x + \log\left(1 - \dfrac{\log x}{x}\right)}\notag\\
&= \lim_{x \to \infty}\frac{\log \log x}{\log x}\cdot\dfrac{1}{1 + \dfrac{\log\left(1 - \dfrac{\log x}{x}\right)}{\log x}}\notag\\
&= 0\notag
\end{align}
A: You are correct to say that $x$ grows faster than $\log(x)$.  And we need neither L'Hospital's Rule nor knowledge of derivatives to prove it.  
In THIS ANSWER, I used only the limit definition of the exponential function and Bernoulli's Inequality to show that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$
for $x>01$.  Now, for any $\alpha$, $\log(x^{\alpha})=\alpha \log(x)$.  Therefore, for $\alpha >0$ we have from $(1)$ we have
$$\frac{x^{\alpha}-1}{\alpha x^{\alpha}}\le \log(x)\le \frac{x^{\alpha}-1}{\alpha} \tag 2$$
Choosing $0<\alpha<1$ in $(2)$, we can assert
$$  x-\frac{x^{\alpha}-1}{\alpha x^{\alpha}} \ge \log(x)\ge x-\frac{x^{\alpha}-1}{\alpha}$$
whereupon applying the squeeze theorem proves the coveted limit
$$\lim_{x\to \infty}x-\log(x)=\infty $$
A: $x-\ln{x}=\ln{e^x}-\ln{x}=\ln{\frac{e^x}{x}}\,,$ and $\frac{e^x}{x}$ is in the form you want.
