First of all, recall the definition of absolute value:
[tex]|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x<0\end{cases}[/tex]
So if x < 4, then x - 4 < 0, so |x - 4| = -(x - 4), and the first case in h(x) reduces to
[tex]\dfrac{|x-4|}{x-4}=\dfrac{-(x-4)}{x-4} = -1[/tex]
Next, in order for h(x) to be continuous at x = 4, the limits from either side of x = 4 must be equal and have the same value as h(x) at x = 4. From the given definition of h(x), we have
[tex]h(4) = 5k-4\cdot4 = 5k-16[/tex]
Compute the one-sided limits:
• From the left:
[tex]\displaystyle \lim_{x\to4^-}h(x) = \lim_{x\to4}\frac{|x-4|}{x-4} = \lim_{x\to4}(-1) = -1[/tex]
• From the right:
[tex]\displaystyle \lim_{x\to4^+}h(x) = \lim_{x\to4}(5k-4x) = 5k-16[/tex]
If the limits are to be equal, then
-1 = 5k - 16
Solve for k :
-1 = 5k - 16
15 = 5k
k = 3
