$\lim\limits_{x \to \infty} \sqrt{\ln(x)} - \sqrt{\ln(x)-\ln(\ln(x))}$

Question

Question:

Find $\lim\limits_{x \to \infty} \sqrt{\ln(x)} - \sqrt{\ln(x)-\ln(\ln(x))}$

For some independent work I would like to analyse this limit. Graphing it, I observe that the function reaches a unique global maximum. Examining the derivative I obtain that this holds when $$x^2 = \ln(x) \exp[(\ln(x)-1)^2]$$

But to obtain a closed form expression for $x$ seems not easy. And even if I prove this, it still doesn't help to compute the limit which is what I am really interested in. I also tried factoring $\sqrt{\ln(x)}$ and using L'Hopital but it didn't help much.

I guess we can also just inspect $\sqrt{y} - \sqrt{y - \ln(y)}$ to understand what is really happening.

Answer 1

Your limit is the same as the limit of $g(x) = \sqrt{x} - \sqrt{x - \ln(x)}$.
Let $x > 1$:
$\begin{align*} g(x) &= g(x) \cdot \frac{\sqrt{x} + \sqrt{x - \ln(x)}}{\sqrt{x} + \sqrt{x - \ln(x)}} = \frac{\sqrt{x}^2 - \sqrt{x - \ln(x)}^2}{\sqrt{x} + \sqrt{x - \ln(x)}} = \frac{x - (x - \ln(x))}{ \sqrt{x} \left(1 + \sqrt{1 - \frac{\ln(x)}{x}}\right)}\\ &= \frac{\ln(x)}{\sqrt{x}}\cdot \frac{1}{ 1 + \sqrt{1 - \frac{\ln(x)}{x}}} \sim \frac{\ln(x)}{2\sqrt{x}} \longrightarrow 0 \end{align*} $

You even have $f(x) = g\circ \ln (x) \sim \frac{\ln(\ln(x))}{\sqrt{\ln(x)}}$, it seems hard to do better, you would have to develop $\sqrt{1 - \frac{\ln(x)}{x}}$ in Taylor series of order $1$.

In similar case where you want to study the behavior of $k(x) = f(g(x)) - f(g(x) - h(x)) $ where $h = o(g)$ in $\infty$, you would like to think about the derivative, indeed, if $f$ is differentiable, you will have $$K(x) \sim f'(g(x)) \cdot h(x)$$ Which here with $f(x) = \sqrt{x}$, $g(x) = \ln(x)$, $h(x) = \ln(\ln(x))$, works perfectly.

Answer 2

Let $\thinspace \ln x=e^{2u}$ with $x>e$ as $u\to \infty $ . Then we have :

$$ \begin{align}\lim_{u\to \infty}e^u-\sqrt {e^{2u}-2u}&=\lim_{u\to \infty}\frac {2u}{e^u+\sqrt {e^{2u}-2u}}\end{align} $$

Since $\thinspace e^{2u}-2u\to \infty$, then notice that :

$$ \begin{align}\frac {2u}{e^u+e^u}&≤\frac {2u}{e^u+\sqrt {e^{2u}-2u}}≤\frac {2u}{e^u}\end{align} $$

Since $\thinspace\lim_{u\to \infty}\dfrac {u}{e^u}=0$, by the Squeeze theorem you are done .

Answer 3

If you want to go beyond the limit itself, let $x=e^t$ and use expansions to show that $$\sqrt{t}-\sqrt{t-\log (t)}=$$ $$\frac{\log \left(\sqrt{t}\right)}{\sqrt{t}}\left( 1+\frac{\log (t)}{4 t}+\frac{\log ^2(t)}{8 t^2}+\frac{5 \log ^3(t)}{64 t^3}+\frac{7 \log ^4(t)}{128 t^4}+O\left(\frac{\log(t)}{t}\right)^5\right)$$ which is a very good approximation.

Limited to first order and back to $x$ $$\sqrt{\log (x)}-\sqrt{\log (x)-\log (\log (x))} \sim \frac{\log \left(\sqrt{\log (x)}\right)}{\sqrt{\log (x)}}~~\to ~0$$

Answer 4

Another way by $y^2=\log x \to \infty$

$$\sqrt{\ln(x)} - \sqrt{\ln(x)-\ln(\ln(x))} =y-\sqrt{y^2-2\ln y}=y\left(1-\sqrt{1-2\frac{\ln y}{y^2}}\right)=$$

$$=\frac{2\frac{\ln y}{y}}{1+\sqrt{1-2\frac{\ln y}{y^2}}} \to \frac{2\cdot 0}{2}=0$$