Question

Calculate each of the limits:

$\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{1+h}}-1}{h}$.

Short answer

The solution for the limit $\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{1+h}}-1}{h}$is$-1$.

Step-by-step solution

Step 1. Given Information.

The limit is,

$\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{1+h}}-1}{h}$.

Step 2. Finding the solution.

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

Inputting $h=0$,

$$\begin{gathered} =\frac{-1}{1+0} \\ =\frac{-1}{1} \\ =-1 \end{gathered}$$