Question 1204215
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        In the post by @MathLover1, this woman made very rude mistakes, which show that
        she not only does not know  Calculus,  but does not know the logarithmic function,  as well.


        Therefore, I will explain the solution from the very beginning.



<pre>
If the problem asks about the limit  at x---> -oo,  then it is clear that the coefficient "a"
in this consideration must be negative - otherwise, logarithm  ln(ax)  is NOT DEFINED.


With negative "a", ln(ax) is defined at negative x, and  we can write  {{{e^ln(ax)}}} = ax.


Then  

    {{{lim(x->-infinity, (e^(ln(ax))*tan((2a)/x)))}}} = {{{lim(x->-infinity, ax*tan((2a)/x))}}} =  {{{2a^2}}} = 8,


which implies {{{a^2}}} = 4,  and since "a" is negative,  a = -2.


So, a = -2 is the only solution to this problem at x ---> -oo, which is exactly opposite to the conclusion by @MathLover1.


If the question is about finding "a" from this equation at x ---> oo,  then  the answer is  a = 2.


Thus, the <U>ANSWER</U> is twofold: if x ---> -oo,  then a = -2.

                             if x --->  oo,   then a = 2.
</pre>

Solved.


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How this woman can call herself &nbsp;" MathLover1 ", &nbsp;making such errors, &nbsp;is a mystery to me.