<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] You looked at each of the three terms separately. You have $|x-1|<\frac{1}{10}$. You should have $|x+1|<\frac{21}{10}$ and you should have $\frac{1}{|x+3|}<\frac{10}{39}$. The critical point for your method is that you have to take the worst case each time. 
The three inequalities then multiply together to give $\left|\frac{x^2-1}{x+3}\right|<\frac{21}{390}<\frac{1}{13}$. 
---------- ASIDE ---------
But note that this method only works because you were given a limit which is substantially larger than the largest possible value of the expression. If you had been asked to show that $$\left|\frac{x^2-1}{x+3}\right|<\frac{1}{19}$$ which is still true for $|x-1|<\frac{1}{10}$ the method would fail.
For a tougher limit the tool of choice is calculus, but one can often solve problems without it. For example in this case we could prove the $\frac{1}{19}$ limit by using the properties of quadratics. 
Note that $x+3$ is positive throughout the range $0.9<x<1$, whereas $x^2-1<0$ for $0.9<x<1$ and $x^2-1>0$ for $1<x<1.1$ so we need to show that $19(1-x^2)<x+3$ for $0.9<x<1$ and $19(x^2-1)<x+3$ for $1<x<1.1$. We have
$$19(1-x^2)-(x+3)=-19x^2-x+16=-19\left(x+\frac{1}{38}\right)^2+16\frac{1}{76}$$ which is strictly decreasing for $x>\frac{1}{38}$. So its largest value in the range $0.9\le x\le 1.1$ is $19(1-0.9^2)-3.9=-0.29<0$ at $x=0.9$.
Similarly $$19(x^2-1)-(x+3)=19\left(x-\frac{1}{38}\right)^2-22\frac{1}{76}$$ which is strictly increasing in the range $1\le x\le 1.1$, so its largest value in the range is $19(1.21^2-1)-4.1=-0.11<0$ at $x=1.1$.
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