Question 1030353
<pre>
{{{matrix(2,1,lim,"x->oo")}}}{{{((x+a)/(x-a))^x}}}{{{""=""}}}{{{e}}}

That will be true if and only if the 
limit of the natural log is 1, so we
reduce the problem to finding "a" if

{{{matrix(2,1,lim,"x->oo")}}}{{{ln(((x+a)/(x-a))^x)}}}{{{""=""}}}{{{1}}}

{{{matrix(2,1,lim,"x->oo")}}}{{{x*ln((x+a)/(x-a))}}}{{{""=""}}}{{{1}}}

{{{matrix(2,1,lim,"x->oo")}}}{{{ln((x+a)/(x-a))/(1/x)}}}{{{""=""}}}{{{1}}}

Since {{{matrix(2,1,lim,"x->oo")}}}{{{((x+a)/(x-a))}}}{{{""=""}}}{{{1}}}
therefore {{{matrix(2,1,lim,"x->oo")}}}{{{ln((x+a)/(x-a))}}}{{{""=""}}}{{{0}}}
And also {{{matrix(2,1,lim,"x->oo")}}}{{{(1/x)}}}{{{""=""}}}{{{0}}}

Both numerator and denominator approach 0, so
we can use L'Hopital's rule

{{{matrix(2,1,lim,"x->oo")}}}{{{((2a/(a^2-x^2))/(-1/x^2)))}}} = 1

which simplifes to

{{{matrix(2,1,lim,"x->oo")}}}{{{(-2ax^2/(a^2-x^2)))}}} = 1

That limit is {{{2a}}}, so we have

{{{2a = 1}}}

{{{a = 1/2}}}

Edwin</pre>