If
[tex]f(x) = \dfrac{2ax+b}{x-1}[/tex]
Note that
[tex]f(2) = \dfrac{4a+b}{2-1} = 4a+b[/tex]
From the given information we have
[tex]\displaystyle \lim_{x\to0} f(x) = \lim_{x\to0} \frac{2ax+b}{x-1} = \frac{0+b}{0-1} = -b = -3 \\\\ ~~~~ \implies b=3[/tex]
and
[tex]\displaystyle \lim_{x\to\infty} f(x) = \lim_{x\to\infty} \frac{2ax+b}{x-1} = \lim_{x\to\infty} \frac{2a+\frac bx}{1-\frac1x} = \frac{2a+0}{1-0} = 2a = 4 \\\\ ~~~~ \implies a=2[/tex]
It follows that
[tex]f(2) = 4\cdot2+3 = 11[/tex]
as required.
