Find the values of $a$ and $b$ such that the function $f$ defined by
$$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{{x - 4}}{{|x - 4|}} + a,}&{{\rm{ if }}x < 4}\\{a + b,}&{{\rm{ if }}x = 4}\\{\frac{{x - 4}}{{|x - 4|}} + b,}&{{\rm{ if }}x > 4}\end{array}} \right.$$.
is a continuous function at $x = 4$}

We have $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{{x - 4}}{{|x - 4|}} + a,}&{{\rm{ if }}x < 4}\\{a + b,}&{{\rm{ if }}x = 4}\\{\frac{{x - 4}}{{|x - 4|}} + b,}&{{\rm{ if }}x > 4}\end{array}} \right.$

At $x = 4,$ ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {4^ - }} \frac{{x - 4}}{{|x - 4|}} + {\rm{a}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{4 - h - 4}}{{|4 - h - 4|}} + a = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{h} + a$
$= - 1 + a$

${\rm{RHL}} = \mathop {\lim }\limits_{x \to {4^ + }} \frac{{x - 4}}{{|x - 4|}} + b$

$= \mathop {\lim }\limits_{h \to 0} \frac{{4 + h - 4}}{{|4 + h - 4|}} + b = \mathop {\lim }\limits_{h \to 0} \frac{h}{h} + b = 1 + b$

$f(4) = a + b \Rightarrow - 1 + a = 1 + b = a + b$

$\Rightarrow$ $- 1 + a = a + b$ and $1 + b = a + b$

Therefore the value of $b = - 1$ and $a = 1$