The function is given as:

$\frac{2x}{3x-1}$

I must prove continuity at $x=1$.

I understand the definitions using the Epsilon Delta proof but my issue comes when I arrive at

$|f(x)-f(1)|=|\frac{1-x}{3x-1}|=\frac{|x-1|}{|3x-1|}$< $ \epsilon$.

I know I must relate that result with $|x-1| < \delta $ but I am stumped on how to bound the denominator of the first result $|3x-1|$

I think I may have figured out my issue, I have been setting $\delta < 1$ but that is not sufficient and I have to go further, I instead and say what if $\delta < 1/3$

In that case I end up with $|3x-1|>1$ so I can go back to my original result to get, $\frac{|x-1|}{|3x-1|} < \frac{|x-1|}{1} = |x-1| < \delta = \epsilon$. So in other words I suppose I should set $\delta = \epsilon$? Or $\delta = min(\epsilon, 1/3)$