$$if \;\;lim(f(x))\Rightarrow 5\;\; with \;\;x\Rightarrow 2$$
This is what I think :/
$$\\\displaystyle\lim_{x\rightarrow2}\;\frac{x^3+ax^2+x+b}{x^2-2}=5\\\\
\displaystyle\lim_{x\rightarrow2}\;\frac{x^3+ax^2+x+b}{x^2-2}*(x-2)=\displaystyle\lim_{x\rightarrow2}\;5(x-2)\\\\
\displaystyle\lim_{x\rightarrow2}\;\frac{x^3+ax^2+x+b}{x+2}=\displaystyle\lim_{x\rightarrow2}\;5x-10\\\\
\displaystyle\lim_{x\rightarrow2}\;x^3+ax^2+x+b=\displaystyle\lim_{x\rightarrow2}\;(5x-10)(x+2)\\\\
8+4a+2+b=0\\\\
4a+b=-10$$
$$if \;\;lim(f(x))\Rightarrow 5\;\; with \;\;x\Rightarrow 2$$
This is what I think :/
$$\\\displaystyle\lim_{x\rightarrow2}\;\frac{x^3+ax^2+x+b}{x^2-2}=5\\\\
\displaystyle\lim_{x\rightarrow2}\;\frac{x^3+ax^2+x+b}{x^2-2}*(x-2)=\displaystyle\lim_{x\rightarrow2}\;5(x-2)\\\\
\displaystyle\lim_{x\rightarrow2}\;\frac{x^3+ax^2+x+b}{x+2}=\displaystyle\lim_{x\rightarrow2}\;5x-10\\\\
\displaystyle\lim_{x\rightarrow2}\;x^3+ax^2+x+b=\displaystyle\lim_{x\rightarrow2}\;(5x-10)(x+2)\\\\
8+4a+2+b=0\\\\
4a+b=-10$$