How to calculate $\lim\limits_{{\rho}\rightarrow 0^+}\frac{\log{(1-(a^{-\rho}+b^{-\rho}-(ab)^{-\rho}))}}{\log{\rho}} $ with $a>1$ and $b>1$?

Question

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.

Answer 1

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 $$\frac{\ln a+\ln b}{\ln a+\ln b}\ne \frac{\ln a}{\ln a}+\frac{\ln b}{\ln b}.$$

Answer 2

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

$$\lim_{\rho\to 0^+}\frac{\log\left(1-a^{-\rho}-b^{-\rho}+(ab)^{-\rho}\right)}{\log(\rho)}=2$$

Answer 3

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 $$\log(a^{-\rho}-1)+\log(b^{-\rho}-1)$$