Left-hand limit of a piecewise function

For $x$ approaching 4 from the left, $x < 4$, so we use the rule $f(x) = \sqrt{4-x}$ for all such values. This is the correct branch to substitute into as $x \to 4^-$.

As $x \to 4^-$, the quantity $4 - x \to 0^+$, so $\sqrt{4-x} \to \sqrt{0} = 0$. Therefore $\displaystyle\lim_{x \to 4^-} f(x) = 0$.

For $x$ approaching 4 from the right, $x \ge 4$, so we use the rule $f(x) = x - 4$ for all such values. This is the correct branch to substitute into as $x \to 4^+$.

As $x \to 4^+$, we compute $x - 4 \to 4 - 4 = 0$. Therefore $\displaystyle\lim_{x \to 4^+} f(x) = 0$.

From the previous parts, $\displaystyle\lim_{x \to 4^-} f(x) = 0$ and $\displaystyle\lim_{x \to 4^+} f(x) = 0$. Since both one-sided limits exist and are equal to each other, the two-sided limit exists.

Therefore $\displaystyle\lim_{x \to 4} f(x) = 0$, the common value of the one-sided limits. Note this holds regardless of the actual value $f(4) = 0$ computed from the formula, since limits depend only on nearby behavior.