By using the definition of limit only, prove that $\lim_{x\rightarrow 0} \frac1{3x+1} = 1$ 
By using the definition of a limit only, prove that
$\lim_{x\rightarrow 0} \dfrac{1}{3x+1} = 1$

We need to find $$0<\left|x\right|<\delta\quad\implies\quad\left|\dfrac{1}{3x+1}-1\right|<\epsilon.$$
I have simplified $\left|\dfrac{1}{3x+1}-1\right|$ down to $\left|\dfrac{-3x}{3x+1}\right|$
Then since ${x\rightarrow 0}$ we can assume $-1<x<1$ then $-2<3x+1<4$ which implies $$\left|\dfrac{1}{3x+1}-1\right|=\left|\dfrac{-3x}{3x+1}\right|<\left|\dfrac{-3x}{4}\right|<\left|\dfrac{-3\delta}{4}\right|=\epsilon$$
No sure if the solution is correct

Note that 
$$\left|\frac{1}{1+3x}-1\right|=\left|\frac{3x}{1+3x}\right| \tag 1$$
Now, we restrict $x$ such that $x\in [-1/4,1/4]$.  And with this restriction, it is easy to see that $1/4/ \le 1+3x$.  Using this in $(1)$ reveals that 
$$\left|\frac{1}{1+3x}-1\right|\le 12|x|\tag 2$$
Finally, given any $\epsilon>0$, 
$$\left|\frac{1}{1+3x}-1\right|<\epsilon$$
whenever $|x|<\delta =\min\left(\frac14,\frac{\epsilon}{12}\right)$.
