Solve this

Solution:

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___.