We're given
[tex]\displaystyle\lim_{x\to2}\sqrt{\frac{f(x)^2-8x+3}{x+1}}=9[/tex]
The function [tex]\sqrt x[/tex] is continuous for all [tex]x>0[/tex]. If a function [tex]g(x)[/tex] is continuous, then
[tex]\displaystyle\lim_{x\to c}g(h(x))=g\left(\lim_{x\to c}h(x)\right)[/tex]
This allows us to pass the limit through the square root:
[tex]\displaystyle\sqrt{\lim_{x\to2}\frac{f(x)^2-8x+3}{x+1}}[/tex]
The limit of a quotient is the quotient of limits (provided the limit of the denominator is not 0):
[tex]\sqrt{\dfrac{\lim\limits_{x\to2}(f(x)^2-8x+3)}{\lim\limits_{x\to2}(x+1)}}[/tex]
In the numerator, we can distribute the limit as
[tex]\displaystyle\lim_{x\to2}(f(x)^2-8x+3)=\left(\lim_{x\to2}f(x)\right)^2+\lim_{x\to2}(3-8x)[/tex]
So we end up with
[tex]\sqrt{\dfrac{\left(\lim\limits_{x\to2}f(x)\right)^2+\lim\limits_{x\to2}(3-8x)}{\lim\limits_{x\to2}(x+1)}}=\sqrt{\dfrac{\left(\lim\limits_{x\to2}f(x)\right)^2-13}3}=9[/tex]
Then we just solve for the desired limit:
[tex]\dfrac{\left(\lim\limits_{x\to2}f(x)\right)^2-13}3=81[/tex]
[tex]\left(\lim\limits_{x\to2}f(x)\right)^2-13=243[/tex]
[tex]\left(\lim\limits_{x\to2}f(x)\right)^2=256[/tex]
[tex]\implies\lim\limits_{x\to2}f(x)=\pm16[/tex]
Obviously the limit can't have two values, so one of these is not right, but only the positive value is one of the answer choices, so the limit is 16.
