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Compute $$\lim_{x \to {0_+}} {\ln (x \ln a) \ln \left({{\ln (ax)}\over{\ln({x\over a})}}\right)}$$ where $a>1$

I am trying to get to a result withou using any advanced methods or even things such as l'Hospital's rule etc..

I got to a phase where I took the limit of the first logarithm which we can see tends to $0$ from rewriting it as $\ln a^x$. Then I wanted to make some adjustments to the second part of the expression and I got to the stage where I have the limit of $$(\ln a^2) \left({x-a\over a}\right)$$ That wont give me exact result but I should be able to justify that it the expression is defined and by multiplying it with the first limit which is $0$, the result should also be $0$.

Can somebody please tell me how correct or wrong I am? Thanks.

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  • $\begingroup$ It's interesting; playing around with graphs seems to imply the answer is just $2\ln a$, unless I entered it in incorrectly. $\endgroup$ – Brian Tung Dec 15 '15 at 22:21
  • $\begingroup$ The limit of the first logarithm is $-\infty$. $\endgroup$ – Bernard Dec 15 '15 at 22:38
  • $\begingroup$ yes, I accidentaly skipped the first logarithm :( $\endgroup$ – user298682 Dec 15 '15 at 22:50
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Let $b = \ln a, y = -\ln x$. Then the limit in question is

\begin{align} L & = \lim_{y \to \infty} (-y+\ln b) \ln\left(\frac{-y+b}{-y-b}\right) \\ & = \lim_{y \to \infty} (-y+\ln b) \ln\left(\frac{y-b}{y+b}\right) \end{align}

If this limit exists, then

\begin{align} e^L & = \lim_{y \to \infty} \left(\frac{y-b}{y+b}\right)^{-y+\ln b} \\ & = \lim_{y \to \infty} \left(1-\frac{2b}{y+b}\right)^{-y+\ln b} \\ & = e^{2b} \end{align}

So, $L = 2b = 2\ln a$.

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Rewrite the expression as $$\ln x\cdot\ln\Bigl(1+ \frac{\ln x}{\ln a}\Bigr)-\ln x\cdot\ln\Bigl(1-\frac{\ln x}{\ln a}\Bigr)=\frac{\ln\Bigl(1+ \cfrac{\ln a}{\ln x}\Bigr)}{\ln x}-\frac{\ln\Bigl(1-\cfrac{\ln a}{\ln x}\Bigr)}{\ln x}$$ Now a standard limit is $\;\displaystyle\lim_{u\to 0}\frac{\ln(1+cu)}u=c$, hence $$\lim_{x\to 0^+}{\ln (x \ln a)\ln \biggl({{\ln (ax)}\over{\ln({x\over a})}}\biggr)}=2\ln a.$$

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