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Show that $f(x)=1/x^2$ is not uniformly continuous one the set $(0,1]$

Using the Sequential Criterion for Nonuniform Continuity - which states that

a function $f:A \rightarrow $ R fails to be uniformly continuous on A iff there exists a particular $\epsilon_0$>0 and two sequences ($x_n$) and ($y_n$) in A, satisfying $|x_n -y_n| \rightarrow 0$, but $|f(x_n) - f(y_n)|\ge \epsilon_0$

I would say:

Take ($x_n$) = $\frac{1}{n}$ and ($y_n$)=$\frac{1}{n^2}$. Obviously $|\frac{1}{n} - \frac{1}{n^2}| \rightarrow 0$, but $|f(x_n) - f(y_n)| = |n^2 - n^4| \ge 12 $, for example for n $\ge$ 2

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This CW answer is intended to remove this question from the Unanswered queue.


There is nothing wrong with your argument involving the Sequential Criterion.

As pointed out in the comments, you could also have approached this by:

  • exhibiting a Cauchy sequence $(x_n)_n$ such that $(f(x_n))_n$ is not Cauchy;
  • showing that the range of $f$ is not bounded, while its domain is.