solve the following. 
If
  $$\lim_{x \to 0^+} \ln(x\ln a)\ln\bigg(\frac{\ln(ax)}{\ln(\frac{x}{a})}\bigg)=6$$
  then find the value of $a$.

 A: We have that
$$\ln(x\ln a)\ln\left(\frac{\ln ax}{\ln\left(\frac{x}{a}\right)}\right)=$$
$$=\left(\ln x+\ln \ln a\right)\ln\left(\frac{\ln x+\ln a}{\ln x-\ln a}\right)=$$
$$=\left(\ln x+\ln \ln a\right)\ln\left(1+\frac{2\ln a}{\ln x-\ln a}\right)=$$
$$=2\ln a\frac{ \ln x+\ln \ln a}{\ln x-\ln a}\frac{\ln\left(1+\frac{2\ln a}{\ln x-\ln a}\right)}{\frac{2\ln a}{\ln x-\ln a}}\to 2\ln a=6 \implies a=e^3$$
indeed by standard limits, since $\log x \to -\infty$, we have that


*

*$u=\frac{2\ln a}{\ln x-\ln a}\to 0\implies\frac{\ln(1+u)}{u}\to 1$


and 


*

*$\frac{\ln x+\ln \ln a}{\ln x-\ln a}=\frac{1+\frac{\ln \ln a}{\ln x}}{1-\frac{\ln a}{\ln x}}\to 1$.

A: Using L'Hopital's rule:
$$\lim_{x \to 0^+} \ln(x\ln a)\ln\bigg(\frac{\ln(ax)}{\ln(\frac{x}{a})}\bigg)=\lim_{x \to 0^+} \frac{\ln(\ln a+\ln x)-\ln(\ln x-\ln a)}{\frac{1}{\ln x+\ln \ln a}}=\\
\lim_{x \to 0^+} \frac{\frac{\frac 1x}{\ln a+\ln x}-\frac{\frac 1x}{\ln x-\ln a}}{-\frac{\frac 1x}{(\ln x+\ln \ln a)^2}}=\lim_{x \to 0^+}\frac{2(\ln x+\ln \ln a)^2\ln a}{\ln^2x -\ln^2 a}=\\
\lim_{x \to 0^+} \frac{4(\ln x+\ln \ln a)\cdot\frac 1x\cdot \ln a}{2\ln x\cdot \frac 1x}=\lim_{x \to 0^+}\frac{2\cdot \frac 1x\cdot \ln a}{\frac 1x}=2\ln a=6 \Rightarrow a=e^3.$$
A: It sometimes helps to clear out some notational clutter with judicious changes of variable. In this case if we write $x=e^u$ (with $u\to-\infty$ as $x\to0^+$) and $a=e^b$, then we have
$$\ln(x\ln a)\ln\left(\ln(ax)\over\ln(x/a)\right)=(u+\ln b)\ln\left(u+b\over u-b\right)=(u+\ln b)\ln\left(1+{2b\over u-b}\right)$$
We might as well also write $v=u-b$, which gives us
$$\ln(x\ln a)\ln\left(\ln(ax)\over\ln(x/a)\right)=v\ln\left(1+{2b\over v}\right)+(b+\ln b)\ln\left(1+{2b\over v}\right)$$
It's clear that $(b+\ln b)\ln(1+2b/v)\to0$ as $v\to-\infty$, since $1+2b/v\to1$ and $\ln1=0$. (Note, there'd be a problem if $b=0$, but it's clear from the $\ln a$ in the original expression that we cannot allow $a=e^0=1$.)  The other limit is routine: If you like, one more change of variable gives
$$\lim_{v\to-\infty}v\ln\left(1+{2b\over v}\right)=2b\lim_{w\to0}{\ln(1+w)\over w}=2b$$
Finally, from $2b=6$ we see $a=e^3$.
