<s>[INST] Prove: if $|x-1|<\frac{1}{10}$ so $\frac{|x^2-1|}{|x+3|}<\frac{1}{13}$ 
Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$

$$|x-1|<\frac{1}{10}$$
$$ -\frac{1}{10}<x-1<\frac{1}{10}$$ 
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
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 $|x+3|<\frac{41}{10}$?
 [/INST] Easier method: use the inequalities
$$|a+b|\le|a|+|b|\quad\hbox{and}\quad |a+b|\ge|a|-|b|\ .$$
If
$$|x-1|<\frac1{10}$$
then
$$|x+1|=|(x-1)+2|\le|x-1|+2<\frac{21}{10}$$
and
$$|x+3|=|4+(x-1)|\ge4-|x-1|>\frac{39}{10}\ .$$
Therefore
$$\frac{|x^2-1|}{|x+3|}=|x-1|\frac{|x+1|}{|x+3|}<\frac1{10}\frac{21/10}{39/10}=\frac{21}{390}<\frac{30}{390}=\frac1{13}\ .$$
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