{\displaystyle {\begin{aligned}{\text{Shannon: }}&I^{\text{S}}(p)=\sum _{i}p_{i}\ln p_{i}\\{\text{Renyi: }}&I_{q}^{\text{R}}(p)={\frac {1}{q-1}}\ln \left(\sum _{i}p_{i}^{q}\right)\\{\text{Tsallis: }}&I_{q}^{\text{T}}(p)={\frac {1}{1-q}}\left(1-\sum _{i}p_{i}^{q}\right)\end{aligned}}}
Wir wissen: {\displaystyle \lim _{x\rightarrow 0}\ln(x+1)=x}
{\displaystyle I_{q}^{\text{R}}(p)\rightarrow I^{\text{S}}(p):}
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{\displaystyle {\begin{aligned}I_{1}^{\text{R}}(p)&:=\lim _{q\rightarrow 1}I_{q}^{\text{R}}(p)\\&=\lim _{q\rightarrow 1}{\frac {1}{q-1}}\ln \left(\sum _{i}p_{i}^{q}\right)\\&=\lim _{q\rightarrow 1}{\frac {1}{q-1}}\underbrace {\ln \left(\underbrace {\sum _{i}p_{i}^{q}} _{\rightarrow 1}-\underbrace {\sum _{i}p_{i}} _{=1}+1\right)} _{\rightarrow \sum _{i}p_{i}^{q}-\sum _{i}p_{i}}\\&=\lim _{q\rightarrow 1}{\frac {1}{q-1}}\left(\sum _{i}p_{i}^{q}-\sum _{i}p_{i}\right)\\&=\lim _{q\rightarrow 1}{\frac {1}{q-1}}\sum _{i}\left(p_{i}^{q}-p_{i}\right)\\&=\lim _{q\rightarrow 1}{\frac {1}{q-1}}\sum _{i}p_{i}\underbrace {\left(p_{i}^{q-1}-1\right)} _{\rightarrow \ln p_{i}^{q-1}}\\&=\lim _{q\rightarrow 1}{\frac {1}{q-1}}\sum _{i}p_{i}\ln p_{i}^{q-1}\\&=\lim _{q\rightarrow 1}{\frac {1}{q-1}}\sum _{i}p_{i}(q-1)\ln p_{i}\\&=\lim _{q\rightarrow 1}{\frac {q-1}{q-1}}\sum _{i}p_{i}\ln p_{i}\\&=\sum _{i}p_{i}\ln p_{i}\\&=I^{\text{S}}(p)\end{aligned}}}
{\displaystyle I_{q}^{\text{T}}(p)\rightarrow I^{\text{S}}(p):}
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{\displaystyle {\begin{aligned}I_{1}^{\text{T}}(p)&:=\lim _{q\rightarrow 1}I_{q}^{\text{T}}(p)\\&=\lim _{q\rightarrow 1}{\frac {1}{1-q}}\left(1-\sum _{i}p_{i}^{q}\right)\\&=\lim _{q\rightarrow 1}{\frac {1}{q-1}}\underbrace {\left(\underbrace {\sum _{i}p_{i}^{q}} _{\rightarrow 1}-1\right)} _{\rightarrow \ln \sum _{i}p_{i}^{q}-1+1}\\&=\lim _{q\rightarrow 1}{\frac {1}{q-1}}\ln \sum _{i}p_{i}^{q}\\&=\lim _{q\rightarrow 1}I_{q}^{\text{R}}(p)\\&=I^{\text{S}}(p)\end{aligned}}}