calculus - By using the definition of limit only, prove that $lim_{xrightarrow 0} frac1{3x+1} = 1$

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 $-1No sure if the solution is correct

Answer

If we focus on $-1

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 $$12\delta < \epsilon$$