Answer
$f(2)=3$
$f'(2)=-1$
Work Step by Step
We have: $\frac{\lim\limits_{x \to 2}x^{3}f(x)-24}{\lim\limits_{x \to 2}x-2}=28$
$\because$ The denominator's limit is 0, and $f$ is differentiable at $x = 2$, the numerator's limit must also be $0$.
$\implies \lim\limits_{x \to 2}x^{3}f(x)-24=0 \implies2^{3}f(2)=24 \implies f(2)=3$
Finding $f'(2)$:
Differentiating both numerator and denominator,
($\because$ According to L'Hopital's rule, $\lim\limits_{x\to c}\frac{f(x)}{g(x)} = \lim\limits_{x\to c}\frac{f'(x)}{g'(x)}$)
$\implies \lim\limits_{x\to2}\frac{x^{3}f'(x)+3x^{2}f(x)}{1}=28$
Substituting $x=2$,
$\implies 8f'(2)+12f(2)=28. $
But, $f(2)=3$ (check above)
$\implies 8f'(2)=28-36=-8 \implies f'(2)=-1$
