By definition of the absolute value, you have
[tex]|4x-1|=\begin{cases}4x-1&\text{for }x\ge\frac14\\1-4x&\text{for }x<\frac14\end{cases}[/tex]
For small [tex]h[/tex], you have [tex]4+h\approx4[/tex], so you would write [tex]|4x-1|=4x-1[/tex]. The limit is then
[tex]\displaystyle\lim_{h\to0}\frac{f(4+h)-f(4)}h=\lim_{h\to0}\frac{(4(4+h)-1)-(4(4)-1)}h=\lim_{h\to0}\frac{4h}h=\lim_{h\to0}4=4[/tex]
