Let $f(x)=\{\begin{array}{ll}3x-4,& 0\le x\le 2\\ 2x+l,& 2<x\le 3\end{array}$. If $f$ is continuous at $x=2$, then what is the value of $l$?

For continuity at $x=2$, the Left-Hand Limit (LHL) must equal the Right-Hand Limit (RHL) and the function value $f(2)$.$f(2)=3(2)-4=2$$\text{LHL}={\lim }_{x\to 2^{-}}(3x-4)=2$$\text{RHL}={\lim }_{x\to 2^{+}}(2x+l)=2(2)+l=4+l$ Setting LHL = RHL:$2=4+l⟹l=-2$