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

For $x < 1$, note $x \ne 1$, so we can simplify $\dfrac{x^2-1}{x-1} = \dfrac{(x-1)(x+1)}{x-1} = x+1$. Thus as $x \to 1^-$, $u(x) = x+1 \to 1+1 = 2$. So $\lim_{x\to 1^-} u(x) = 2$.

For $x \ge 1$, $u(x) = x+3$, so as $x \to 1^+$, $u(x) \to 1+3 = 4$. Thus $\lim_{x\to 1^+} u(x) = 4$. Since the left-hand limit ($2$) and right-hand limit ($4$) are not equal, the two-sided limit $\lim_{x\to 1} u(x)$ does not exist.