Question

Find

$\lim_{h \to \0} \frac{ f(4+h)-f(4)}{h}$ if f(x) = absolute value of (4x-1)

Cheers! :)

Answer 1

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

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