Showing 1/x is continuous on (0,1] I am trying to prove that 1/x is continuous on (0,1]. However I am not sure if my way is the correct way, because it seems like I am showing that 1/x is continous everywhere which is not the case.. Here is my try
$|f(x)-f(c)|=|\frac{1}{x}-\frac{1}{c}|=\frac{|x-c|}{xc}$
So given $\epsilon >0 $ we can choose $\delta = \frac{\epsilon}{xc}$
And then $|x-c|< \delta \implies |f(x)-f(c)|<\frac{\epsilon}{xc}xc=\epsilon $
for $x,c \in (0,1]$
 A: I would start by being a little bit more explicit regarding what $c,x$ represent.
$\forall c\in(0,1], \forall \epsilon >0,\exists \delta >0, \forall x\in(0,1]: |c-x|<\delta \implies |f(c) - f(x)|<\epsilon$
Next I would say something like let $\delta<\frac {c}{2}.$ Then $x>\frac {c}{2},$ and $xc > \frac {c^2}{2}$
$|\frac 1c - \frac1x| = |\frac {x-c}{xc}|<\frac{2|x-c|}{c^2}$
Now I see that when $c\in (0,1]$ I have created a bit of redundancy as
$\frac {c^2}{2} < \frac {c}{2}$
When $\delta = \frac{c^2\epsilon}{2}$
$|x-c| < \delta \implies |\frac 1c - \frac1x| = |\frac {x-c}{xc}|<\frac{2|x-c|}{c^2} <\epsilon$
A: Your $$\delta = \frac{\epsilon}{xc}$$ should not depend on your variable $x$.
Also in dividing by $cx$ you are assuming that $c\ne 0$ so you are not proving continuity everywhere which is good.    
A: Can't have your $\delta$ dependent on $x$ as this must be true for all $x$ within $\delta$ of $c$.  The concept "all $x$ within $\delta$ of $c$" would be meaningless if the value of $\delta$ changes for different values of $x$.
So for any $c \in (0,1]$ and any $\epsilon > 0$ we need to find a $\delta$ (based an $\epsilon$ and maybe on $c$; but not on $x$ which we haven't picked yet) so that for any $x\in (0,1]$ where $|x - c| < \delta$ it will have to follow that $|\frac 1x - \frac 1c|< \epsilon$.
If  $|x - c| < \delta$ then $-\delta < x-c < \delta$ and $c - \delta < x < c + \delta$. If we assume that $c-\delta > 0$ (we'll have to deal with $c-\delta \le 0$ later...) then $\frac 1{c+\delta} < \frac 1x < \frac 1{c - \delta}$.
This would imply $\frac 1{c + \delta} - \frac 1c= -(\frac 1c - \frac 1{c+\delta}) < \frac 1x - \frac 1c < \frac 1{c-\delta} - \frac 1c$ and that $|\frac 1x - \frac 1c| < \max (\frac 1c - \frac 1{c+\delta}, \frac 1{c-\delta} - \frac 1c)$.
$\frac 1c - \frac 1{c+\delta} = \frac {\delta}{c(c+\delta)} < \frac {\delta}{c(c-\delta)} = \frac 1{c-\delta} - \frac 1c$ so if we had managed to define $\delta$ is such a way that $\frac 1{c-\delta} - \frac 1c \le \epsilon$ we would be done.
And to do that $\frac 1{c-\delta} - \frac 1c = \frac {\delta}{c(c-\delta)} \le \epsilon$ would mean $\delta \le \epsilon c(c-\delta)\epsilon$ or
$\delta + c\delta*\epsilon \le c^2 \epsilon$ or
$\delta \le \frac {c^2 \epsilon}{1+ c\epsilon}$ will do.
But we have to worry about $\delta < c$ .... but that can't happen as $\delta = \frac {c^2 \epsilon}{1+ c\epsilon}= \frac {c}{\frac 1{c\epsilon} + 1} < c$. 
That's it.  If $\delta = \frac {c}{\frac 1{c\epsilon} + 1}$ then we can prove that if $|x - c| < \delta$ then $|\frac 1x - \frac 1c| < \epsilon$.
