0
$\begingroup$

Let $f(x)=\frac{1}{x-3}$ on $(3,4]$. I need to show that f is continuous in this interval, but not uniformly continuous.

Idea: Since $f'(x)=\frac{-1}{(x-3)^{2}}$, the derivative exists at all points in $(3,4]$ then $f$ is continuous in $(3,4]$ But if we look at uniform continuity, $\lim_{x \to a^{+}}f(x)$ is infinity while $f(3)$ is simply undefined. Is this accurate to say? More worried about the second part.

  • 0
    For continuity, all you really need is that it is a rational function, and those are continuous wherever they are defined.2012-11-28
  • 1
    The derivative exists at all points in (3,4], so as you note, $f$ is continuous there. But is the derivative $f'$ *bounded* on (3,4]? What is the implication for uniform continuity?2012-11-28

2 Answers 2