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 $|x+3|<\frac{41}{10}$?

The gradient of $f(x)=\dfrac{x^2-1}{x+3}$ near $x=1$ is $$f'(x)=1-\dfrac{8}{(x+3)^2},$$which is positive in the range $0.9<x<1.1$; so, in this range, $|f(x)|$ ranges from $0$ (at $x=1$) to $\max\{|f(0.9)|,|f(1.1)|\}=0.0512...<1/13.$
