How can I prove the limit of this function using epsilon-delta?

Question

Let $f: \mathbb{R} \setminus \{\frac{4}{7}\} \mapsto \mathbb{R}$ be defined by $f(x) = \frac{1}{7x-4}$. Using the epsilon-delta definition, show that $\lim_{x\to 1} f(x) = \frac{1}{3}$. So, I know that I need to find $\delta$ such that $0<|x-1|<\delta \implies |\frac{1}{7x-4}-\frac{1}{3}| < \epsilon$. This is what I have so far: $$ |\frac{1}{7x-4}-\frac{1}{3}| < \epsilon \\ -\epsilon < \frac{1}{7x-4}-\frac{1}{3} < \epsilon \\ -\epsilon +\frac{1}{3} < \frac{1}{7x-4} < \epsilon +\frac{1}{3} \\ -\frac{1}{\epsilon}+3 < 7x-4 < \frac{1}{\epsilon} + 3 \\ -\frac{1}{\epsilon}+7 < 7x < \frac{1}{\epsilon} + 7 \\ -\frac{1}{7\epsilon}+1 < x < \frac{1}{7\epsilon} + 1 \\ -\frac{1}{7\epsilon} < x-1 < \frac{1}{7\epsilon} \\ $$ This next part is where I got confused. So, I said pick $\delta=\min(-\frac{1}{\epsilon},\frac{1}{\epsilon})$. I am not sure how to proceed from here. Thanks in advance!