How to use the $\;\epsilon-\delta\;$ definition to show that $\lim\limits_{x \to 1} \frac{1}{3x+1}=\frac14$

Question

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 $\;\delta$.

Answer 1

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 $\delta=\min\left(1,\frac{4\epsilon}3\right)$.

Answer 2

Let $\delta >0$, being small.

Let $0<\lvert x-1\rvert<\delta.$

$$\left\lvert \frac{1}{3x+1} -\frac{1}{4}\right\rvert =\left\lvert \frac{3-3x}{12x+4}\right\rvert = \frac{3}{4}\left\lvert \frac{1-x}{3x+1}\right\rvert \le\\ \le\frac{3}{4} \frac{\delta}{2-3\delta}\le\frac{3}{4}\delta \to0$$ as $\delta \to 0,$ where we have used the fact that $\lvert x \rvert>1-\delta$ and $\lvert 3x+1\rvert \ge \lvert 3x\rvert -1 \gt 2-3\delta.$