Question 1204233
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The natural log and base 'e' are inverses of each other.
e^(ln(ax)) = ax


As *[tex \Large \text{x} \to -\infty], then ln(ax) approaches positive infinity when a < 0.


The equation
*[tex \Large \lim_{\text{x} \to -\infty} e^{\ln(a\text{x})}\tan\left(\frac{2a}{\text{x}}\right) = 8]
is the same as
*[tex \Large \lim_{\text{x} \to -\infty} a\text{x}\tan\left(\frac{2a}{\text{x}}\right) = 8]


The {{{ax}}} portion approaches positive infinity as *[tex \Large \text{x} \to -\infty] when a < 0. 


The *[tex \Large \tan\left(\frac{2a}{\text{x}}\right)] portion approaches tan(0) = 0 as *[tex \Large \text{x} \to -\infty].
This is because the 2a/x part approaches 0 as x approaches positive infinity.


This will mean
*[tex \Large \lim_{\text{x} \to -\infty} a\text{x}\tan\left(\frac{2a}{\text{x}}\right) = 8]
turns into
*[tex \Large \infty*0 = 8]
The left hand side is one of the <a href="https://en.wikipedia.org/wiki/Indeterminate_form">indeterminate forms</a> in calculus.


What you'll need to do is rewrite one of the expressions so that you have a ratio of two functions.


This is one rewrite we could do
{{{ax*tan((2a)/x)}}} ---> {{{(tan((2a)/x))/(1/(ax))}}}
The new equivalent expression is of the form P/Q where
{{{P = tan((2a)/x)}}} and {{{Q = 1/(ax)}}}


From here, use L'Hopital's rule to apply the derivative to functions P and Q.
Then apply the limit to see if you get another indeterminate form or not. 
If so, then apply L'Hopital's rule again.
If not, then you'll be able to solve for 'a'.


I'll let the student take over from here.
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