calculus - 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 :)

Answer

$x-\ln{x}=\ln{e^x}-\ln{x}=\ln{\frac{e^x}{x}}\,,$ and $\frac{e^x}{x}$ is in the form you want.