Is there some trick/formula to finding your delta value when proving continuity and/or uniform continuity? This seems to be very hit or miss for me, and I find myself often finding having a lot of trouble with it.

For instance, if we have $f(x) = \frac{1}{x^2 + 1}$, and want to prove it's uniformly continuous from $\mathbb{R} \rightarrow \mathbb{R}$, apparently the delta value that's useful is $\epsilon / 2$. But, I just cannot see how to get this value.

Any help would be greatly appreciated. Thank you.

  • 1
    $\begingroup$ Most of these tricks come from solving the problem in reverse. Assume that $|f(x) -f(y)|<\epsilon$. Now, plug in for $f(x),f(y)$, and try to figure out how you need to bound the term $|x-y|$ to get that $|f(x) - f(y)|<\epsilon$ $\endgroup$ – rubikscube09 May 7 '18 at 18:50

For a differentiable function like that, one trick is to see if you can bound the derivative as less than some constant $C$. Then you can find a good $\delta$ since

$$|f(x) - f(y)| \leq C |x - y|$$

So, if you want $ |f(x) - f(y) | < \epsilon$, then you can just choose $\delta < \frac{\epsilon}{C}$.

  • $\begingroup$ How is the derivative of $\frac{1}{x^2 + 1}$ bounded by 2? $\endgroup$ – Axioms May 7 '18 at 19:18
  • $\begingroup$ @Axioms The derivative is $\frac{-2x}{(1+x^2)^2}$. We can bound its magnitude by $2$ by splitting into cases, one where $0 \leq |x| \leq 1$ and one for $|x| > 1$. $\endgroup$ – Michael Biro May 7 '18 at 21:01

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