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# Need help proving continuity for $\frac{2x}{3x-1}$ at $x=1$ using Epsilon-Delta

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 $$\delta = min(\epsilon, 1/3)$$