One-sided limits and existence of a two-sided limit

For $x < 3$, we have $x - 3 < 0$, so $|x-3| = -(x-3)$, giving $\dfrac{|x-3|}{x-3} = -1$. Thus $h_2(x) = -1 + 1 = 0$ for all $x$ near $3$ from the left, so $\lim_{x\to 3^-} h_2(x) = 0$.

For $x > 3$, we have $x - 3 > 0$, so $|x-3| = x-3$, giving $\dfrac{|x-3|}{x-3} = 1$. Thus $h_2(x) = 1 + 1 = 2$ for all $x$ near $3$ from the right, so $\lim_{x\to 3^+} h_2(x) = 2$.

Since the left-hand limit ($0$) and the right-hand limit ($2$) are not equal, the two-sided limit $\lim_{x\to 3} h_2(x)$ does not exist. Note that the value $h_2(3) = 5$ plays no role in this conclusion, since limits depend only on behavior near $x=3$, not at $x=3$.