If f(x) = begin{cases} frac{x-2}{|x-2|}+a & , | vedclass

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(C) For $f(x)$ to be continuous at $x=2$, the left-hand limit $(LHL)$, right-hand limit $(RHL)$, and the value of the function at $x=2$ must be equal.

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(C) For $f(x)$ to be continuous at $x=2$, the left-hand limit $(LHL)$, right-hand limit $(RHL)$, and the value of the function at $x=2$ must be equal. $1$. Calculate $LHL$: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^-} (\frac{x-2}{|x-2|} + a)$. Since $x $2$. Calculate $RHL$: $\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^+} (\frac{x-2}{|x-2|} + b)$. Since $x > 2$, $|x-2| = (x-2)$, so $\frac{x-2}{x-2} + b = 1 + b$. $3$. Value at $x=2$: $f(2) = a + b$. Equating these: $-1 + a = 1 + b = a + b$. From $-1 + a = a + b$, we get $b = -1$. From $1 + b = a + b$, we get $a = 1$. Therefore, $a + b = 1 + (-1) = 0$.

If $f(x) = \begin{cases} \frac{x-2}{|x-2|}+a & , x<2 \\ a+b & , x=2 \\ \frac{x-2}{|x-2|}+b & , x>2 \end{cases}$ is continuous at $x=2$, then $a+b=$

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