How to use the $\;\epsilon-\delta\;$ definition to show that $\lim\limits_{x \to 1} \frac{1}{3x+1}=\frac14$ I have tried evaluating the $\;\delta\;$ part of the definition. So far I have reached the following point:
$0<|x-1|<\delta\;$ implies $\;\left|\frac{1}{3x+1}-\frac14\right|<\epsilon$
which I simplified to be $\;3|1-x|\left|\frac{1}{12x+4}\right|<\epsilon\;$,
but I am unsure where to go from here to obtain $\;\delta$.

If $0<|x-1|<\delta$ and $\delta<1$, then $x\in(0,2)$ so $12x+4\in(4,28)$, so $\left|\frac{1}{12x+4}\right|<\frac14$.
So take $\delta=\min\left(1,\frac{4\epsilon}3\right)$.
