If $f(x)=\left\{\begin{array}{cl}\frac{x-4}{|x-4|}+a, & x<4 \\ a+b, & x=4 \\ \frac{x-4}{|x-4|}+b, & x>4\end{array}\right.$. Then $f(x)$ is continuous at $x=4$, then $a+b=$
The function $f(x)=\left\{\begin{array}{ll}\frac{x-4}{|x-4|}+a, & x<4 \\ a+b, & x=4 \\ \frac{x-4}{|x-4|}+b, & x>4\end{array}\right.$ is continuous at $x=4$.
$\therefore f(4)=\lim _{x \rightarrow 4} f(x)$
$\Rightarrow f(4)=\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)$
$|x-4|= \begin{cases}-(x-4), & x<4 \\ x-4, & x \geq 4\end{cases}$
Now,
$f(4)=a+b$ ...(2)
$\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4}\left[\frac{x-4}{-(x-4)}+a\right]=\lim _{x \rightarrow 4} \frac{x-4}{-(x-4)}+\lim _{x \rightarrow 4} a=-1+a$ ......(3)
$\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4}\left[\frac{x-4}{(x-4)}+b\right]=\lim _{x \rightarrow 4} \frac{x-4}{(x-4)}+\lim _{x \rightarrow 4} b=1+b$ ....(4)
From (1), (2), (3) and (4), we get
$a+b=-1+a=1+b$
So,
$a+b=-1+a$
$\Rightarrow b=-1$
Also,
$a+b=1+b$
$\Rightarrow a=1$
$\therefore a+b=1+(-1)=0$
Thus, the value of a + b is 0.
If $f(x)=\left\{\begin{array}{cl}\frac{x-4}{|x-4|}+a, & x<4 \\ a+b, & x=4 \\ \frac{x-4}{|x-4|}+b, & x>4\end{array}\right.$. Then $f(x)$ is continuous at $x=4$, then $a+b=$ ___0___.