<s>[INST] 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
 [/INST] If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $\frac{1}{3x+1}$is unbounded.
Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < \frac14$. Hence $\delta < \frac14$.
if $-\frac14 < x < \frac14$, $$-\frac34+1 < 3x+1< \frac34+1$$.
$$\frac14 < 3x+1< \frac74$$
$$\left|\frac{1}{3x+4} \right| < 4$$
Hence $$12\delta < \epsilon$$
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