Find the values of a and b such that the function f defined by
$f(x)= \begin{cases}\frac{x-4}{|x-4|}+a, & \text { if } x<4 \\ a+b & , \text { if } x=4 \\ \frac{x-4}{|x-4|}+b, & \text { if } x>4\end{cases}$
is a continuous function at x = 4.
Finding the left hand and right hand limit for the given function, we have
$\lim _{x \rightarrow 4} f(x)=\frac{x-4}{|x-4|}+a$
$=\lim _{h \rightarrow 0} \frac{4-h-4}{|4-h-4|}+a=\lim _{h \rightarrow 0} \frac{-h}{h}+a=-1+a$
$\lim _{x \rightarrow 4} f(x)=a+b$
$\lim _{x \rightarrow 4^{-}} f(x)=\frac{x-4}{|x-4|}+b$
$=\lim _{h \rightarrow 0} \frac{4+h-4}{|4+h-4|}+b=\lim _{h \rightarrow 0} \frac{h}{h}+b=1+b$
As the function is continuous at $x=4$.
$\therefore \quad \lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4} f(x)=\lim _{x \rightarrow 4^{-}} f(x)$
So, -1 + a = a + b = 1 + b
-1 + a = a + b and 1 + b = a + b
We get, b = -1 and 1 + -1 = a + -1 ⇒ a = 1
Therefore, the value of a = 1 and b = -1
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