Question 1206594
<pre>
If I'm reading what you wrote correctly, it's this

If {{{matrix(1,2,lim, ("f(x)"/(x-2)))}}}{{{""=""}}}{{{5}}}, {{{f(x)/(2x-1)=15/4}}}, then {{{

matrix(1,2,
matrix(2,1,lim,"x->-1"),("f(x)"/(x+1)))="?"}}}

The first part makes no sense because there is nothing given for x to approach
in the limit.  So I can only ignore that, and consider it as

If {{{f(x)/(2x-1)=15/4}}}, then {{{matrix(1,2,
matrix(2,1,lim,"x->-1"),("f(x)"/(x+1)))="?"}}}


{{{f(x)/(2x-1)=15/4}}}

{{{4f(x)=15(2x-1)}}}

{{{f(x)=expr(15/4)(2x-1)}}}

{{{matrix(1,2,
matrix(2,1,lim,"x->-1"),("f(x)"/(x+1)))}}}{{{""=""}}}{{{matrix(1,2,
matrix(2,1,lim,"x->-1"),(expr(15/4)(2x-1)/(x+1)))^""))}}}{{{""=""}}}{{{matrix(1,2,
expr(15/4)matrix(2,1,lim,"x->-1"),((2x-1)/(x+1)))}}}

Make a change of variable x+1=y, x=y-1

{{{matrix(1,2,
expr(15/4)matrix(2,1,lim,"y-1->-1"),((2(y-1)-1)/((y-1)+1)))}}}{{{""=""}}}

{{{matrix(1,2,
expr(15/4)matrix(2,1,lim,"y->0"),((2y-2-1)/(y-1+1)))}}}{{{""=""}}}

{{{matrix(1,2,
expr(15/4)matrix(2,1,lim,"y->0"),((2y-3)/(y)))}}}{{{""=""}}}

{{{matrix(1,2,
expr(15/4)matrix(2,1,lim,"y->0"),((2y-3)/(y)))}}}{{{""=""}}}{{{expr(15/4)*2}}}{{{""=""}}}{{{15/2}}}

Edwin</pre>