Q.For what values of $a$ and $b$ is the function $f(x) = \dfrac{x^2-4}{x-2}$, for $x < 2$, $= ax^2-bx+3$, for $2 \le x < 3$, $= 2x-a+b$, for $x \ge 3$, continuous for every $x$ on $\mathbb{R}$?

$f(x)=\dfrac{x^2-4}{x-2}=x+2$ (for $x\ne2$, hence for all $x<2$) for $x<2$; $=ax^2-bx+3$ for $2\le x<3$; $=2x-a+b$ for $x\ge3$. Continuity is needed at $x=2$ and $x=3$.

At $x=2$: Left-hand limit: $\displaystyle\lim_{x\to2^-}(x+2)=4$. Value: $f(2)=a(4)-b(2)+3=4a-2b+3$ (middle piece, since domain is $2\le x<3$). Setting equal: $4a-2b+3=4\Rightarrow4a-2b=1.\quad(i)$

At $x=3$: Left-hand limit using the middle piece: $\displaystyle\lim_{x\to3^-}(ax^2-bx+3)=9a-3b+3$. Value/right-hand side: $f(3)=2(3)-a+b=6-a+b$ (third piece, domain $x\ge3$). Setting equal: $9a-3b+3=6-a+b\Rightarrow10a-4b=3.\quad(ii)$ …

First simplify the rational first piece to a plain linear function via cancellation, then set up and solve the pair of continuity equ …

Forgetting to simplify $\frac{x^2-4}{x-2}$ to $x+2$ before taking its limit at $x=2$, or misassigning which piece 'owns' each junction point (the middle piece owns bot …