Question

Please help with this limit i don’t know what to do at all!

Answer 1

We're given

$\displaystyle\lim_{x\to2}\sqrt{\frac{f(x)^2-8x+3}{x+1}}=9$

The function $\sqrt x$ is continuous for all $x>0$. If a function $g(x)$ is continuous, then

$\displaystyle\lim_{x\to c}g(h(x))=g\left(\lim_{x\to c}h(x)\right)$

This allows us to pass the limit through the square root:

$\displaystyle\sqrt{\lim_{x\to2}\frac{f(x)^2-8x+3}{x+1}}$

The limit of a quotient is the quotient of limits (provided the limit of the denominator is not 0):

$\sqrt{\dfrac{\lim\limits_{x\to2}(f(x)^2-8x+3)}{\lim\limits_{x\to2}(x+1)}}$

In the numerator, we can distribute the limit as

$\displaystyle\lim_{x\to2}(f(x)^2-8x+3)=\left(\lim_{x\to2}f(x)\right)^2+\lim_{x\to2}(3-8x)$

So we end up with

$\sqrt{\dfrac{\left(\lim\limits_{x\to2}f(x)\right)^2+\lim\limits_{x\to2}(3-8x)}{\lim\limits_{x\to2}(x+1)}}=\sqrt{\dfrac{\left(\lim\limits_{x\to2}f(x)\right)^2-13}3}=9$

Then we just solve for the desired limit:

$\dfrac{\left(\lim\limits_{x\to2}f(x)\right)^2-13}3=81$

$\left(\lim\limits_{x\to2}f(x)\right)^2-13=243$

$\left(\lim\limits_{x\to2}f(x)\right)^2=256$

$\implies\lim\limits_{x\to2}f(x)=\pm16$

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.