Prove the following identities.


$f(x)=\left\{\begin{array}{ll}\frac{|x-4|}{2(x-4)} & \text { if } x \neq 4 \\ 0, & \text { if } x=4\end{array}\right.$ at $\mathrm{x}=4$


Checking the right hand and left hand limits for the given function, we have

$\lim _{x \rightarrow 4} f(x)=\frac{|x-4|}{2(x-4)} \quad\left[\begin{array}{c}\text { for } x<4,|x-4|=-(x-4) \\ \text { for } x>4,|x-4|=(x-4)\end{array}\right]$

$=\lim _{h \rightarrow 0} \frac{-[4-h-4]}{2[4-h-4]}=\lim _{h \rightarrow 0} \frac{h}{-2 h}=-\frac{1}{2}$

$\lim _{x \rightarrow 4^{+}} f(x)=\frac{|x-4|}{2(x-4)}=\lim _{h \rightarrow 0} \frac{[4+h-4]}{2[4+h-4]}=\frac{1}{2}$

$\lim _{x \rightarrow 4} f(x)=0$

$\therefore \quad \lim _{x \rightarrow 4^{-}} f(x) \neq \lim _{x \rightarrow 4^{-}} f(x) \neq \lim _{x \rightarrow 4} f(x)$

Thus, f(x) is discontinuous at x = 4.

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