The right hand limit of the function 

  $f\left(x\right)=\left\{\begin{array}{l}\frac{|x-4|}{x-4},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x\ne 4\\ 0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=4\end{array}\right.$

at x = 4, is 

Solution:

We have,

$\begin{array}{l}\text{ (RHL of }f\left(x\right)\text{ at }x=4\text{ ) }\\ =\underset{x\to 4^{+}}{lim} f\left(x\right)=\underset{h\to 0}{lim} f(4+h)=\underset{h\to 0}{lim} \frac{|4+h-4|}{4+h-4}\\ =\underset{h\to 0}{lim} \frac{\left|h\right|}{h}=\underset{h\to 0}{lim} \frac{h}{h}=\underset{h\to 0}{lim} 1=1.\end{array}$