Originally Posted by

**eskimo343** We have not seen that result yet, so I am trying to prove this __just__ using the definition of uniform continuity. So, I did this:

Let $\displaystyle \delta= (a-1)^2 \cdot \epsilon$. We assume that $\displaystyle |x-y|<\delta$. Then $\displaystyle |\frac{x}{x-1}-\frac{y}{y-1}|$$\displaystyle =\frac{|x(y-1)-y(x-1)|}{|(x-1)(y-1)|} \leq \frac{|x(y-1)-y(x-1)|}{(a-1)^2}$$\displaystyle =\frac{|y-x|}{(a-1)^2}=\frac{|x-y|}{(a-1)^2}

< \frac{\delta}{(a-1)^2}= \frac{(a-1)^2 \cdot \epsilon}{(a-1)^2}= \epsilon.$ So, I am really just wondering if this works to prove uniform continuity on $\displaystyle [a,2]$. Does this show uniform continuity on $\displaystyle [a,2]$ just using the definition of uniform continuity?