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Continuity proof

  1. Aug 1, 2009 #1
    So today I wanted to prove that [itex]x/(x-k)[/itex] is continuous for [itex]x\neq k[/itex]. I have to show that for all [itex]\varepsilon>0[/itex] there exists a [itex]\delta[/itex] such that [itex]\left|x/(x-k)-x_0/(x_0-k)\right|<\varepsilon[/itex] for all [itex]x[/itex] satisfying [itex]|x-x_0|<\delta[/itex]. This is how I did it (a bit long)

    [tex]\begin{align*}\left|\frac{x}{x-k}-\frac{x_{0}}{x_{0}-k}\right| & =\left|\frac{x-x_{0}}{x_{0}-k}+\frac{x}{x-k}-\frac{x}{x_{0}-k}\right|\\
    & =\left|\frac{x-x_{0}}{x_{0}-k}+\frac{x\left((x_{0}-k)-(x-k)\right)}{(x-k)(x_{0}-k)}\right|\\
    & =\left|\frac{x-x_{0}}{x_{0}-k}+\frac{(x-x_{0}+x_{0})\left((x_{0}-k)-(x-k)\right)}{(x-k)(x_{0}-k)}\right|\\
    & \le\left|\frac{x-x_{0}}{x_{0}-k}\right|+\frac{|x-x_{0}|\left|x_{0}-x\right|}{\left|(x-k)(x_{0}-k)\right|}+\frac{|x_{0}|\left|x_{0}-x\right|}{\left|(x-k)(x_{0}-k)\right|}\\
    & =\frac{\delta}{|x_{0}-k|}+\frac{\left(\delta+|x_{0}|\right)\delta}{\left|(x-k)(x_{0}-k)\right|}\\
    & \le\frac{\delta}{|x_{0}-k|}+\frac{\left(\delta+|x_{0}|\right)\delta}{\left||x-x_{0}|-|x_{0}-k|\right||x_{0}-k|}\\
    & =\frac{\delta}{|x_{0}-k|}+\frac{\left(\delta+|x_{0}|\right)\delta}{\left|\delta-|x_{0}-k|\right||x_{0}-k|}<\varepsilon.\end{align*}

    Now at this point, what is the easiest way to argue that a [itex]\delta[/itex] exists? Can this problem be solved with less arithmetic? I guess the idea is that you want an upper limit which consists of [itex]\delta[/itex] and some constants, right?
  2. jcsd
  3. Aug 1, 2009 #2


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    Not sure if this approach is simpler but I would attack the problem in this fashion: First we prove that the functions f(x) = x and g(x) = (x - k)-1 are continuous for all real x with x != k. This is not only simple but also requires virtually no arithmetic - especially if you're familiar with limits. The only proof that is left then is to prove that the product of the functions must also be continuous for all x with x != k.
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