Question
Express $|3 x|<\frac{1}{3}$ in interval form.
| Comparing $|3x| < \frac{1}{3}$ with $|x - a|<δ$ we get |
| $a = 0$ and $δ = \frac{1}{9}$. |
| Interval form $: N(a - δ ; a + δ)$ |
| Putting $a = 0$ and $δ =\frac{1}{9}$. |
| $|3 x|<\frac{1}{3}=\left(0-\frac{1}{9} ; 0+\frac{1}{9}\right)$ |
| $=\left(-\frac{1}{9} ; \frac{1}{9}\right)$ |
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