• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q. 64

Expert-verified
Found in: Page 136

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Calculate each of the limits:$\underset{h\to 0}{\mathrm{lim}}\frac{{\left(2+h\right)}^{2}-{2}^{2}}{h}$.

The solution of the limit $\underset{h\to 0}{\mathrm{lim}}\frac{{\left(2+h\right)}^{2}-{2}^{2}}{h}$ is, $4$.

See the step by step solution

## Step 1. Given Information.

The limit is,

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

## Step 2. Finding the solution.

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

Using ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$,

$=\underset{h\to 0}{\mathrm{lim}}\frac{{h}^{2}+4h+4-4}{h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{h}^{2}+4h}{h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}h+4$

Inputting $h=0$

$=0+4\phantom{\rule{0ex}{0ex}}=4$