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$$\lim_{x\to\ 0^+}(e^x-1)\ln(\ln(1+x))$$

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 ?

Etemon
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    We can't use equivalence in this way, morover $\ln(\ln(1+x)) \to -\infty$ is not equivalent to $(x-1)$. We can proceed by Taylor's series but we need to use remainder terms. In this case, standard limits suffice, we don't need l'Hospital. – user Oct 29 '20 at 15:12

3 Answers3

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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.

Claude Leibovici
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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

user
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  • So we used Auxiliary fractions like $\frac{e^x-1}{x}$ and the fraction $\frac{x}{\ln(1+x)}$ to make sure 2 terms of taylor series is sufficient to conclude the limit of them is $1$. am I right? – Etemon Oct 29 '20 at 15:14
  • And why $\lim_{t\to 0}t\ln t=0$? I am not familiar with that – Etemon Oct 29 '20 at 15:15
  • @soheil The first 2 you have indicated are fundamental standard limits which can be proved without Taylor's or l'Hospital. I'll give a reference for that. – user Oct 29 '20 at 15:15
  • Also $\lim_{t\to 0}t\ln t=0$ is a standard limit which can be proved using $t=\frac 1 y$ as $y \to \infty$ then $$\lim_{t\to 0}t\ln t=\lim_{y\to \infty}-\frac{\ln y}y =0$$ – user Oct 29 '20 at 15:17
  • For $\lim_{t\to 0}t\ln t$ isn't possible to use equivalence and get $t(t-1)=0\times(-1)=0$? – Etemon Oct 29 '20 at 15:23
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    @soheil Note that $\ln t \to -\infty$ which is not equivalent to $(t-1) \to -1$. This is a wrong equivalence. In any way, I suggest warmly to do not use equivalence to solve limit. Often this leads to wrong results. It can be useful for a first evaluation but it is not good for a rigorous proof. – user Oct 29 '20 at 15:26
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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

$$(e^x-1)\ln(\ln(1+x))\sim x\log(x)\to0.$$