Right-hand limit of a piecewise absolute value ratio

When $x \to 4^+$, we are considering values of $x$ slightly larger than $4$, so $x - 4 > 0$. In this case, the absolute value simplifies as $|x-4| = x-4$, since the quantity inside is already positive.

Substituting this into $f(x)$ gives $f(x) = \frac{x-4}{x-4} = 1$ for all $x$ near $4$ with $x > 4$. Therefore $\lim_{x \to 4^+} \frac{|x-4|}{x-4} = 1$.

When $x \to 4^-$, we are considering values of $x$ slightly smaller than $4$, so $x - 4 < 0$. In this case, the absolute value simplifies as $|x-4| = -(x-4)$, since the quantity inside is negative.

Substituting this into $f(x)$ gives $f(x) = \frac{-(x-4)}{x-4} = -1$ for all $x$ near $4$ with $x < 4$. Therefore $\lim_{x \to 4^-} \frac{|x-4|}{x-4} = -1$.

We found that $\lim_{x \to 4^+} f(x) = 1$ and $\lim_{x \to 4^-} f(x) = -1$. Since these two one-sided limits are not equal to each other, the function does not approach a single common value as $x$ approaches $4$ from both sides.

Because the two-sided limit exists if and only if both one-sided limits exist and agree, and here they disagree ($1 \neq -1$), we conclude that $\lim_{x \to 4} \frac{|x-4|}{x-4}$ does not exist.