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Let be $n \in \mathbb{N}$ arbitrary but fixed, $\sum\limits_{i=1}^n p_i =1$ and $\forall ~ 1\leq i \leq n$ we assume: $x_i \in \mathbb{R}$.

What is the limit of

$\lim\limits_{r\to 0}~\frac{1}{r}\ln\left(1+r\sum\limits_{i=1}^n p_i \ln(x_i)+ \omicron(r)\right)$, where $\lim\limits_{r\to 0}\frac{\omicron(r)}{r}=0$?

I tried some manipulations and the theorem of L'Hospital but it only got worse...

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  • $\begingroup$ This is not well stated. Is $n$ fixed? What are the $x_i?$ What are the $p_i?$ $\endgroup$ – zhw. Jun 2 at 22:21
  • $\begingroup$ Thanks for the remark. I have edited the question. $\endgroup$ – Philipp Jun 2 at 22:27
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Let $$S(r)=r\sum_{i=1}^np_i\ln(x_i)$$

we have $$\lim_{r\to 0}(S(r)+o(r))=0$$

thus

$$\color{red}{\lim_{r\to 0}\frac{\ln\Bigl(1+S(r)+o(r)\Bigr)}{S(r)+o(r)}=1}$$

therefore, the limit is $$\lim_{r\to0}\frac{S(r)+o(r)}{r}=\sum_{i=1}^np_i\ln(x_i)$$ because

$$\frac{\ln(1+S(r)+o(r))}{r}=$$ $$\color{red}{\frac{\ln(1+S(r)+o(r))}{S(r)+o(r)}}\frac{S(r)+o(r)}{r}.$$

and $$\frac{S(r)}{r}+\frac{o(r)}{r}=$$ $$\sum_{i=1}^np_i\ln(x_i)+\frac{o(r)}{r}$$

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  • $\begingroup$ Can you explain to me the step that follows "the limit is ..." a little bit more? Why can we conclude the limit of $\lim_{r\to0}\frac{S(r)+o(r)}{r}$ from $\lim_{r\to 0}\frac{\ln\Bigl(1+S(r)+o(r)\Bigr)}{S(r)+o(r)}=1$ ? $\endgroup$ – Philipp Jun 2 at 22:16
  • $\begingroup$ @Philipp I just added some lines in the end. hope will help. $\endgroup$ – hamam_Abdallah Jun 2 at 22:32
  • $\begingroup$ Thank you for the further explanations :) $\endgroup$ – Philipp Jun 2 at 23:49
  • $\begingroup$ Do you know if it is possible to show the statement with the theorem of L'Hospital? The problem is that when applying this theorem there appears a term $\omicron'(r)$ and I am not sure what happens to this term when $r \to 0$. $\endgroup$ – Philipp Jun 3 at 0:01
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Hint

Remembering that $$ \lim_{r \to 0} \frac{\ln(1+r)}{r} = 1 $$

we get $$ \begin{split} \lim_{r \to 0}\frac{\ln\left(1+r\sum\limits_{i=1}^n p_i \ln(x_i)+ \omicron(r)\right)}{r} &= \lim_{r \to 0}\frac{\ln\left(1+r\sum\limits_{i=1}^n p_i \ln(x_i)+ \omicron(r)\right)}{\sum\limits_{i=1}^n p_i \ln(x_i)+ \omicron(r)}\cdot \frac{r\sum\limits_{i=1}^n p_i \ln(x_i)+ \omicron(r)}{r} \end{split} $$

Can you continue from here?

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