If $f(x)= \begin{cases}\frac{x-4}{|x-4|}+a, & \text { if } \quad x<4 \\ a+b \quad, & \text { if } \quad x=4 \text { is continuous at } x=4, \text { find } a, b . \\ \frac{x-4}{|x-4|}+b, & \text { if } \quad x>4\end{cases}$
Given:
$f(x)=\left\{\begin{array}{c}\frac{x-4}{|x-4|}+a, \text { if } \mathrm{x}<4 \\ a+b, \text { if } \mathrm{x}=4 \\ \frac{x-4}{|x-4|}+b, \text { if } \mathrm{x}>4\end{array}\right.$
We observe
$(\mathrm{LHL}$ at $x=4)=\lim _{x \rightarrow 4^{-}} f(x)=\lim _{h \rightarrow 0} f(4-h)$
$=\lim _{h \rightarrow 0}\left(\frac{4-h-4}{|4-h-4|}+a\right)=\lim _{h \rightarrow 0}\left(\frac{-h}{|-h|}+a\right)=a-1$
$(\mathrm{RHL}$ at $x=4)=\lim _{x \rightarrow 4^{+}} f(x)=\lim _{h \rightarrow 0} f(4+h)$
$=\lim _{h \rightarrow 0}\left(\frac{4+h-4}{|4+h-4|}+b\right)=\lim _{h \rightarrow 0}\left(\frac{h}{|h|}+b\right)=b+1$
And $f(4)=a+b$
If $f(x)$ is continuous at $x=4$, then
$\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)=f(4)$
$\Rightarrow a-1=b+1=a+b$
$\Rightarrow a-1=a+b, b+1=a+b$
$\Rightarrow b=-1, a=1$