$ = \mathop {\lim }\limits_{h \to 0} - \frac{h}{h} + a = a - 1.$
$ = \mathop {\lim }\limits_{x \to 4 + } f(x) = \mathop {\lim }\limits_{h \to 0} \,\,f(4 + h) = \mathop {\lim }\limits_{h \to 0} \,\frac{{4 + h - 4}}{{|4 + h - 4|}} + b = b + 1$
and $f(4) = a + b$
Since $f(x)$ is continuous at $x = 4$
Therefore $\mathop {\lim }\limits_{x \to 4 - } f(x) = f(4) = \mathop {\lim }\limits_{x \to 4 + } f(x)$
$ \Rightarrow \,\,a - 1 = a + b = b + 1\,\, \Rightarrow \,\,b = - 1$ and $a = 1.$
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