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 […]