How to calculate $\lim\limits_{{\rho}\rightarrow 0^+}\frac{\log{(1-(a^{-\rho}+b^{-\rho}-(ab)^{-\rho}))}}{\log{\rho}} $ with $a>1$ and $b>1$? How to calculate this question?
$$\lim\limits_{{\rho}\rightarrow 0^+}\frac{\log{(1-(a^{-\rho}+b^{-\rho}-(ab)^{-\rho}))}}{\log{\rho}} ,$$
where $a>1$ and $b>1$.
Thank you everyone.
 A: First, note that you have $(1-a^{-\rho})(1-b^{-\rho})$
inside the logarithm in the denominator, and that
$$
\frac{d}{d\rho}a^{-\rho}=
\frac{d}{d\rho}e^{-\rho\ln a}=
-e^{-\rho\ln a}\cdot\ln a=
-a^{-\rho}\ln a.
$$
So
$$
\begin{align}L&=
\lim_{\rho\to0^+}
\frac{\log{(1-(a^{-\rho}+b^{-\rho}-(ab)^{-\rho}))}}{\log{\rho}}
\\&=\lim_{\rho\to0^+}
\frac{\log(1-e^{-\rho\ln a})+\log(1-e^{-\rho\ln b})}{\log{\rho}}
\\&\stackrel*=\lim_{\rho\to0^+}
\rho\left[
\frac{a^{-\rho}\ln a}{1-a^{-\rho}}+
\frac{b^{-\rho}\ln b}{1-b^{-\rho}}
\right]
\\&=\lim_{\rho\to0^+}
\frac{\rho\ln a}{a^\rho-1}+
\frac{\rho\ln b}{b^\rho-1}
\\&\stackrel*=\lim_{\rho\to0^+}
\frac{\ln a}{a^\rho\ln a}+
\frac{\ln b}{b^\rho\ln b}
\\&=2
\end{align}
$$
where we have used L'Hopital's Rule ($*$) twice and algebra on the remaining lines. Finally, note that while my derivation assumes that
$\log=\ln$ are both base $e$, the answer is the same if $\log$
is base $10$ because
$$\frac{d}{dx}\log_{10} x
=\frac{d}{dx}\frac{\ln x}{\ln 10}
=\frac1{x\ln 10},$$
and the two factors of $\ln 10$ would cancel out after the first
application of L'Hopital's Rule.
The original post suggested the answer should be $1$.
Such a proposed result might be the consequence
of erroneously arriving at
$$\frac{\ln a+\ln b}{\ln a+\ln b}\ne
\frac{\ln a}{\ln a}+\frac{\ln b}{\ln b}.$$
A: Using the expansions
$$e^z=1+z+\frac12 z^2+O(z^3)$$
and 
$$\log(1+x)=O(x)$$
we can write
$$\begin{align}
\log\left(1-a^{-\rho}-b^{-\rho}+(ab)^{-\rho}\right)&=\log\left(1-e^{-\log(a)\rho}-e^{-\log(b)\rho}+e^{-\log(ab)\rho}\right)\\\\
&=\log\left(\log(a)\log(b)\rho^2+O(\rho^3)\right)\\\\
&=\log(\log(a)\log(b)\rho^2)+O(\rho)\\\\
&=\log(\log(a)\log(b))+2\log(\rho)+O(\rho)
\end{align}$$
Therefore, we have
$$\lim_{\rho\to 0^+}\frac{\log\left(1-a^{-\rho}-b^{-\rho}+(ab)^{-\rho}\right)}{\log(\rho)}=2$$
A: Factoring the argument of $\log$ we can see that the numerator can be written as $$\log(1-a^{-\rho})+\log(1-b^{-\rho})$$ and note that we have $$\lim_{\rho\to 0^{+}}\frac{a^{\rho}-1}{\rho}=\log a$$ which implies that $$\lim_{\rho\to 0^{+}}\frac{1-a^{-\rho}}{\rho\log a}=1$$ We can now write $$\frac{\log(1-a^{-\rho})}{\log\rho} = \dfrac{\log\left(\dfrac{1-a^{-\rho}}{\rho\log a}\right)+\log\rho +\log\log a}{\log\rho} = 1 +\dfrac{\log\left(\dfrac{1-a^{-\rho}}{\rho\log a}\right) +\log\log a}{\log\rho}$$ so that this expression tends to $1$ as $\rho\to 0^{+}$. Similarly $\dfrac{\log(1-b^{-\rho})}{\log\rho} \to 1$ as $\rho\to 0^{+}$. Hence the desired limit is $1+1=2$. The answer is valid even when $0<a,b<1$ but then the numerator needs to be split as $$\log(a^{-\rho}-1)+\log(b^{-\rho}-1)$$
