Evaluating $\lim_{x\to1^+}\ln x\ln(\ln x)$ without L'hospital

Question

I want to evaluate the following limit :$$L=\lim_{x\to1^+}\ln x\ln(\ln x)$$

My efforts:

$$L =\lim_{x\to 1^+} \frac{\ln(\ln x)}{\frac{1}{\ln x}}$$

By L'hospital rule, clearly :

$$L = \lim_{x\to1^+}(-\ln x)=0$$

But I am looking into any other way to evaluate it. Thanks !

Answer 1

Let $x=e^t$, then the limit is: $$\lim_{t\to 0^+} t \log t=0 $$ which can be proven without L'Hopital's rule as done here.

Answer 2

$$L=\lim_{x\to 1}\ln x\ln(\ln x)=\lim_{x\to 1}\ln\left((\ln x)^{\ln x}\right)=\ln \lim_{x\to 1}\left((\ln x)^{\ln x}\right).$$ So we have to compute $$\lim_{x\to 1}\left((\ln x)^{\ln x}\right).$$ Take $t=\ln x$, so the limit converts to $$\lim_{t\to 0^+}t^t,$$ whose limit is $1$, this have been answered here many times. So $$L=\ln 1=0.$$ The fact that $$\lim_{t\to 0^+}t^t,$$ is basically that $x\ln x\to 0$ as $x\to 0^+$. To show this without using L'hopital check this $\lim_{x\to0^{+}} x \ln x$ without l'Hopital's rule

Answer 3

By another way, let $x=e^{e^y}\to 1^+$ with $y\to -\infty$ then

$$\lim_{x\to1^+}\ln x\ln(\ln x)=\lim_{y\to-\infty }te^t=$$

which is a well known standard limit (exponential vs power) which goes to $0$.

Answer 4

When $x\to 1$ we have $l = \ln(x)\sim x-1$, so: $$l=\lim_{x\to 1^+}\ln(x)\cdot\ln(\ln(x))\sim \lim_{x\to 1^+}(x-1)\cdot\ln(x-1)$$

Let $t=x-1$, thus: $$l=\lim_{t\to 0^+} t\cdot\ln(t)=\lim_{s\to +\infty}-\frac{\ln(s)}{s}=0$$ for $s=1/t$.