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Verified Answer :

LammettHash

By definition of the absolute value, you have

\(|4x-1|=\begin{cases}4x-1&\text{for }x\ge\frac14\\1-4x&\text{for }x<\frac14\end{cases}\)

For small \(h\), you have \(4+h\approx4\), so you would write \(|4x-1|=4x-1\). The limit is then

\(\displaystyle\lim_{h\to0}\frac{f(4+h)-f(4)}h=\lim_{h\to0}\frac{(4(4+h)-1)-(4(4)-1)}h=\lim_{h\to0}\frac{4h}h=\lim_{h\to0}4=4\)