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Q. 65

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:$\underset{h\to 0}{\mathrm{lim}}\frac{{\left(1+h\right)}^{3}-{1}^{3}}{h}$.

The solution for the limit $\underset{h\to 0}{\mathrm{lim}}\frac{{\left(1+h\right)}^{3}-{1}^{3}}{h}$ is, $1$.

See the step by step solution

## Step 1. Given Information.

The limit is,

$\underset{h\to 0}{\mathrm{lim}}\frac{{\left(1+h\right)}^{3}-{1}^{3}}{h}$.

## Step 2. Finding the solution.

$\underset{h\to 0}{\mathrm{lim}}\frac{{\left(1+h\right)}^{3}-{1}^{3}}{h}$

Using ${\left(a+b\right)}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}$,

$=\underset{h\to 0}{\mathrm{lim}}\frac{1+{h}^{3}+3h+3{h}^{2}-1}{h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{h}^{3}+3{h}^{2}+h}{h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left({h}^{2}+3h+1\right)$

Inputting $h=0$,

$={0}^{2}+3\left(0\right)+1\phantom{\rule{0ex}{0ex}}=0+0+1\phantom{\rule{0ex}{0ex}}=1$