Show that $\displaystyle f(x)=\frac 1{x-1}$ is not uniformly continuous on $\displaystyle (1,6)$.
Take $\displaystyle \epsilon=10$.
Suppose there exists $\displaystyle 0<\delta<1$ satisfyng the definition of uniform convergence.
Choose $\displaystyle x=1+\delta,\;y=1+\delta/11$.
Prove that $\displaystyle |x-y|<\delta$ and $\displaystyle |f(x)-f(y)|>10$ (absurd)
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