Q. 45 - 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:

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

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

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