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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 :)

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    $\begingroup$ The derivative of $x - \ln x$ is $1 - 1/x$, which is always positive for $x > 1$, so your quantity grows without bound :) $\endgroup$ – amcerbu Mar 13 '16 at 0:20
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$x-\ln{x}=\ln{e^x}-\ln{x}=\ln{\frac{e^x}{x}}\,,$ and $\frac{e^x}{x}$ is in the form you want.

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    $\begingroup$ That is really elegant, thanks! $\endgroup$ – Chewers Jingoist Mar 13 '16 at 18:54
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    $\begingroup$ @ChewersJingoist: This is why it is a really good idea not to pick best answers so quickly. Although I agree that this is a nicer answer than Eff's, it's kind of rude to remove someone's best answer status while the question is so young. $\endgroup$ – Eric Stucky Mar 13 '16 at 21:27
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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.

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  • $\begingroup$ The function $f(x)=1-1/x$ also satisfies $f'(x)>0$ for all $x>1$, but this doesn't mean the difference between $1$ and $1/x$ increases without bound. Your argument does carry though with the additional observation that $\lim_{x\to\infty} f'(x)\neq 0$. $\endgroup$ – Eric Stucky Mar 13 '16 at 0:34
  • $\begingroup$ @EricStucky Yes, you're right. In fact, $f'(x)$ approaches 1. $\endgroup$ – Eff Mar 13 '16 at 12:18
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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 $$\frac{\ln\ln x}{\ln(x-\ln x)}\to 0 \enspace\text{as}\enspace x\to +\infty$$

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    $\begingroup$ Doesn't using $x-\ln x \sim_\infty x$ beg the question? $\endgroup$ – Eric Stucky Mar 13 '16 at 1:28
  • $\begingroup$ I don't think so. The equivalence stems from the fact that $x-\ln x=x+o(x)$, which is an elementary result (without the notation) from high school. $\endgroup$ – Bernard Mar 13 '16 at 1:46
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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}

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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

$$\lim_{x\to \infty}x-\log(x)=\infty $$

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