0

Verify that $f(x)=\frac{1}{x}-\frac{1}{x_0}$ is continuous for every $x_0\neq 0$.

$f(0)$ is not defined. So the function is discontinuous at $0$.

Let $c\in \mathbb{R}\setminus \lbrace 0 \rbrace$, we have that

$\lim_{x \to c} f(x)=\lim_{x \to c} (\frac{1}{x}-\frac{1}{x_0})= \frac{1}{c}-\frac{1}{x_0}=f(c)$

Then $f$ continuous on $\mathbb{R}\setminus \lbrace 0 \rbrace $.

Is that true, please?

GNUSupporter 8964民主女神 地下教會
  • 17,998
Dima
  • 2,501
  • 1
    I suspect that your argument is valid, although I would have also gone to the trouble of using an epsilon/delta argument to demonstrate that $;\lim_{x\to c} \frac{1}{x} = \frac{1}{c}.$ – user2661923 Feb 16 '19 at 17:56
  • 1
    Regarding “so the function is discontinuous at 0”: https://math.stackexchange.com/questions/1087623/is-function-f-mathbb-c-0-rightarrow-mathbb-c-prescribed-by-z-rightarrow – Hans Lundmark Feb 16 '19 at 18:20
  • 1
    when you say $\lim_{x\to c} (\frac 1x - \frac 1{x_0}) = \frac 1c - \frac 1{x_0}$ why do you know you can just plug in $x = c$ into the equation? Isn't that precisely what you need to prove? – fleablood Feb 16 '19 at 19:29
  • Thanks everybody, I will try to prove that. – Dima Feb 16 '19 at 19:35

0 Answers0