Is it ok to use equivalence here where we have $\lim0\times(-\infty)$? $$\lim_{x\to\ 0^+}(e^x-1)\ln(\ln(1+x))$$
Here to evaluate the limit I used the equivalence $\ln(u+1)\sim u$ twice  :
$$\lim_{x\to\ 0^+}(e^x-1)\ln(\ln(1+x))=\lim_{x\to 0^+}(e^x-1)\times(x-1)=(1-1)\times(0-1)=0$$
But the book I am reading write the expression as $\cfrac{\ln(\ln(1+x))}{\cfrac1{e^x-1}}$ then applied L'Hopital Rule and at the end it get $0$ too after relatively long calculation.
I wonder is my approach valid ?
 A: You can even go beyond the limit composing Taylor series one piece at the time
$$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$
$$\log(\log(1+x))=\log (x)-\frac{x}{2}+O\left(x^2\right)$$
$$e^x-1=x+\frac{x^2}{2}+O\left(x^3\right)$$
$$(e^x-1)\log(\log(1+x))=\left(x+\frac{x^2}{2}+O\left(x^3\right) \right) \left(\log (x)-\frac{x}{2}+O\left(x^2\right) \right) $$
$$(e^x-1)\log(\log(1+x))=x \log (x)+\frac{1}{2} x^2 (\log (x)-1)+O\left(x^3\right)$$ which for sure shows the limit but also how it is approached.
A: No we can't use equivalence in this way, we can proceed by standard limits using that
$$(e^x-1)\ln(\ln(1+x))=\frac{e^x-1}x \cdot x\ln(\ln(1+x)) \to 1 \cdot 0=0$$
indeed
$$x\ln(\ln(1+x))=\ln(1+x)\cdot\ln(\ln(1+x))\cdot \frac{x}{\ln(1+x)} \to 0 \cdot 1 =0$$
since for $t\to 0$ we have that $t\ln t \to 0$.
Refer to the related

*

*How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

*How can I deduce that $\lim\limits_{x\to0} \frac{\ln(1+x)}x=1$ without Taylor series or L'Hospital's rule?
A: There is no room to apply it twice.
$$\log(\log(1+x))\sim\log(x)$$ and $x$ is not $1+x$ !
Tinkering with $x=1+(x-1)$ does not work because $x-1$ is not close to $0$.

On the opposite, $$e^x-1\sim x$$ will work and finally
$$(e^x-1)\ln(\ln(1+x))\sim x\log(x)\to0.$$
