Let f(x) = begin{cases} frac{x - 4}{|x - 4|} | vedclass

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(D) For $f(x)$ to be continuous at $x = 4$,the left-hand limit,right-hand limit,and the value of the function at $x = 4$ must be equal.Left-hand limit

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$a = 1, b = -1$

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(D) For $f(x)$ to be continuous at $x = 4$,the left-hand limit,right-hand limit,and the value of the function at $x = 4$ must be equal.Left-hand limit $(LHL)$: $\lim_{x 	o 4^-} f(x) = \lim_{h 	o 0} f(4 - h) = \lim_{h 	o 0} (\frac{4 - h - 4}{|4 - h - 4|} + a) = \lim_{h 	o 0} (\frac{-h}{|-h|} + a) = -1 + a$.Right-hand limit $(RHL)$: $\lim_{x 	o 4^+} f(x) = \lim_{h 	o 0} f(4 + h) = \lim_{h 	o 0} (\frac{4 + h - 4}{|4 + h - 4|} + b) = \lim_{h 	o 0} (\frac{h}{|h|} + b) = 1 + b$.Value of the function: $f(4) = a + b$.For continuity: $a - 1 = a + b = b + 1$.From $a - 1 = a + b$,we get $b = -1$.From $a + b = b + 1$,we get $a = 1$.Thus,$a = 1$ and $b = -1$.

Let $f(x) = \begin{cases} \frac{x - 4}{|x - 4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x - 4}{|x - 4|} + b, & x > 4 \end{cases}$. Then $f(x)$ is continuous at $x = 4$ when

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