Question

Calculate each of the limits

$\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(2+h)}^2}}-1}{h}$

Short answer

$\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(2+h)}^2}-1}}{h}=-1$

Step-by-step solution

Step 1. Given information is

$\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(2+h)}^2}-1}}{h}$

Step 2. Solving and using the formula (a+b)2=a2+2ab+b2

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(2+h)}^2}-1}}{h}=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4-{(2+h)}^2}{{(2+h)}^2}}}{h} \\ \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(2+h)}^2}-1}}{h}=\underset{h\to 0}{\lim }\frac{4-(4+4h+h^2)}{h{(2+h)}^2} \\ \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(2+h)}^2}-1}}{h}=\underset{h\to 0}{\lim }\frac{-4h-h^2}{h{(2+h)}^2} \\ \underset{h\to 0}{\lim }\frac{4-(4+4h+h^2)}{h{(2+h)}^2}=\underset{h\to 0}{\lim }\frac{-h(4+h)}{h{(2+h)}^2} \\ \underset{h\to 0}{\lim }\frac{-4h-h^2}{h{(2+h)}^2}=\underset{h\to 0}{\lim }\frac{-(4+h)}{{(2+h)}^2} \end{gathered}$$

Step 3. Solving the limit we get,

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{-4h-h^2}{h{(2+h)}^2}=\underset{h\to 0}{\lim }\frac{-(4+h)}{{(2+h)}^2} \\ \underset{h\to 0}{\lim }\frac{-4h-h^2}{h{(2+h)}^2}=\underset{h\to 0}{\lim }\frac{-(4+0)}{{(2+0)}^2} \\ \underset{h\to 0}{\lim }\frac{-4h-h^2}{h{(2+h)}^2}=\frac{-4}{4} \\ \underset{h\to 0}{\lim }\frac{-4h-h^2}{h{(2+h)}^2}=-1 \end{gathered}$$