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 $|x+3|<\frac{41}{10}$?
 A: Your method (what you are trying to do) is correct, but you have a few errors.

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

This is wrong. It should be
$$|x+1|\lt \frac{21}{10}$$

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

This is wrong. It should be
$$|x+3|\lt\frac{41}{10}$$
but we use
$$|x+3|\color{red}{\gt} \frac{39}{10}\iff \frac{1}{|x+3|}\lt \frac{10}{39}$$
since $|x+3|$ is in the denominator.
Therefore, we get
$$\frac{|x^2-1|}{|x+3|}\lt\frac{1}{10}\cdot\frac{21}{10}\cdot\frac{10}{39}=\frac{7}{130}\lt \frac{10}{130}=\frac{1}{13}$$
A: 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
$$\frac{|x^2-1|}{|x+3|}=|x-1|\frac{|x+1|}{|x+3|}<\frac1{10}\frac{21/10}{39/10}=\frac{21}{390}<\frac{30}{390}=\frac1{13}\ .$$
A: You are very close but you have actually made two mistakes that have 'cancelled' each other out to get the right result. 
Like you pointed out the inequality you used was wrong. In fact, you have the reverse of this inequality. However, this is exactly what you need since you have to remember that dividing reverses the direction of the inequality as well. Hence, the rest of the proof is correct. 
A: 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 $\max\{|f(0.9)|,|f(1.1)|\}=0.0512...<1/13.$
A: 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 $x=1.1$.
