Left-hand limit as x approaches 4

For $x$ values approaching 4 from the left (like 3.9, 3.99, 3.999), we have $0 \le x < 4$, so $f(x) = \sqrt{x}$ applies. As $x \to 4^-$, $\sqrt{x}$ approaches $\sqrt{4} = 2$.

Therefore $\lim_{x \to 4^-} f(x) = 2$, since the square root function is continuous and $\sqrt{x} \to \sqrt{4}$ as $x \to 4$ from values less than 4.

For $x$ values approaching 4 from the right (like 4.1, 4.01, 4.001), we have $x \ge 4$, so $f(x) = 6 - x$ applies. As $x \to 4^+$, $6 - x$ approaches $6 - 4 = 2$.

Therefore $\lim_{x \to 4^+} f(x) = 2$, since the linear function $6-x$ is continuous everywhere and evaluates to 2 at $x=4$.

We found $\lim_{x \to 4^-} f(x) = 2$ and $\lim_{x \to 4^+} f(x) = 2$. Since both one-sided limits exist and are equal to the same value.

By the rule that a two-sided limit exists if and only if the one-sided limits exist and agree, we conclude $\lim_{x \to 4} f(x) = 2$.