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> Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$
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$$|x-1|<\frac{1}{10}$$

$$ -\frac{1}{10}<x-1<\frac{1}{10}$$ 

$$ \frac{19}{10}<x+1<\frac{21}{10}$$

$$|x+1|<\frac{19}{10}$$

Adding 4 to both sides of $$ -\frac{1}{10}<x-1<\frac{1}{10}$$ gives:

$$\frac{39}{10}<x+3<\frac{41}{10}$$

$$|x+3|<\frac{39}{10}$$

Plugging those results in $$\frac{|x-1|\*|x+1|}{|x+3|}<\frac{1}{13}$$
We get: $$\frac{\frac{1}{10}\*\frac{19}{10}}{\frac{39}{10}}<\frac{1}{13}$$

$$\frac{19}{390}<\frac{1}{13}$$ Which is true, is this proof is valid as I took the smallest intervals, like $|x+3|<\frac{39}{10}$ and not $|x+3|<\frac{41}{10}$?