Proving Continuity of $\frac{x}{x-k}$ for $x\neq k$

[daudaudaudau]

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

$$\begin{align*}\left|\frac{x}{x-k}-\frac{x_{0}}{x_{0}-k}\right| & =\left|\frac{x-x_{0}}{x_{0}-k}+\frac{x}{x-k}-\frac{x}{x_{0}-k}\right|\\<br /> & =\left|\frac{x-x_{0}}{x_{0}-k}+\frac{x\left((x_{0}-k)-(x-k)\right)}{(x-k)(x_{0}-k)}\right|\\<br /> & =\left|\frac{x-x_{0}}{x_{0}-k}+\frac{(x-x_{0}+x_{0})\left((x_{0}-k)-(x-k)\right)}{(x-k)(x_{0}-k)}\right|\\<br /> & \le\left|\frac{x-x_{0}}{x_{0}-k}\right|+\frac{|x-x_{0}|\left|x_{0}-x\right|}{\left|(x-k)(x_{0}-k)\right|}+\frac{|x_{0}|\left|x_{0}-x\right|}{\left|(x-k)(x_{0}-k)\right|}\\<br /> & =\frac{\delta}{|x_{0}-k|}+\frac{\left(\delta+|x_{0}|\right)\delta}{\left|(x-k)(x_{0}-k)\right|}\\<br /> & \le\frac{\delta}{|x_{0}-k|}+\frac{\left(\delta+|x_{0}|\right)\delta}{\left||x-x_{0}|-|x_{0}-k|\right||x_{0}-k|}\\<br /> & =\frac{\delta}{|x_{0}-k|}+\frac{\left(\delta+|x_{0}|\right)\delta}{\left|\delta-|x_{0}-k|\right||x_{0}-k|}<\varepsilon.\end{align*}$$

Now at this point, what is the easiest way to argue that a $\delta$ exists? Can this problem be solved with less arithmetic? I guess the idea is that you want an upper limit which consists of $\delta$ and some constants, right?

 

[jgens]

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.