Q. 46 - Calculus

Question

Calculate each of the limits in Exercises $29-70$.

$\underset{h\to 0}{\lim }\frac{{\left(-1+h\right)}^2-1}{h}$.

Step-by-Step Solution

Answer

The value of $\underset{h\to 0}{\lim }\frac{{\left(-1+h\right)}^2-1}{h}$ is, $-2$.

Step 1 . Given information

$\underset{h\to 0}{\lim }\frac{{\left(-1+h\right)}^2-1}{h}$.

Step 2 . The limit is calculated as below:

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

                              $=\underset{h\to 0}{\lim }(-2)+\underset{h\to 0}{\lim }\left(h\right)$ [Sum rule]

                               $$\begin{gathered} =-2+0 \\ =-2 \end{gathered}$$