Question

Calculate each of the limits

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

Short answer

$\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(1+h)}^2}-4}}{h}=-8$

Step-by-step solution

Step 1. The given information is

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

Step 2. Solving we get,

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(1+h)}^2}-4}}{h}=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4-4{(1+h)}^2}{{(1+h)}^2}}}{h} \\ \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(1+h)}^2}-4}}{h}=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4-4(1+2h+h^2)}{{(1+h)}^2}}}{h} \\ \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(1+h)}^2}-4}}{h}=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4-4-8h-4h^2}{{(1+h)}^2}}}{h} \\ \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(1+h)}^2}-4}}{h}=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{-4h(2+h)}{{(1+h)}^2}}}{h} \\ \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(1+h)}^2}-4}}{h}=\underset{h\to 0}{\lim }\frac{-8-4h}{{(1+h)}^2} \end{gathered}$$

Step 3. Solving limits we get,

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(1+h)}^2}-4}}{h}=\underset{h\to 0}{\lim }\frac{-8-4h}{{(1+h)}^2} \\ \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(1+h)}^2}-4}}{h}=\underset{h\to 0}{\lim }\frac{-8-4\left(0\right)}{{(1+0)}^2} \\ \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{4}{{(1+h)}^2}-4}}{h}=-8 \end{gathered}$$