# Thread: Proving continuity using the epsilon-delta criterion - question 2

1. ## Proving continuity using the epsilon-delta criterion - question 2

I figure the best way to see if i'm understanding this is to try it on another example:

Show that the function $\displaystyle f0, \infty) \rightarrow \mathbb{R}$ given by $\displaystyle f(x)=\frac{1}{x}$ is continuous using the usual $\displaystyle \epsilon-\delta$ definition.
(this is the following question).

$\displaystyle \forall \epsilon>0 \exists \delta_{\epsilon}: \forall x \in E: |x-x_0|< \delta_{\epsilon} \Rightarrow |f(x)-f(x_0)|< \epsilon$

In this case, $\displaystyle |f(x)-f(x_0)| < \epsilon$ is $\displaystyle |\frac{1}{x}-\frac{1}{x_0}| < \epsilon$.

Starting from $\displaystyle |x-x_0|<\delta_{\epsilon} \leq 1$

$\displaystyle \frac{|x-x_0|}{|-xx_0|} < \frac{\delta_{\epsilon}}{|-xx_0|}$ where $\displaystyle x,x_0 \neq 0$ (which is acceptable since $\displaystyle \frac{1}{x} \neq 0$)

$\displaystyle \frac{|x_0-x|}{|xx_0|}< \frac{\delta_{\epsilon}}{|xx_0|}$

$\displaystyle |\frac{1}{x}-\frac{1}{x_0}| < \frac{\delta_{\epsilon}}{|xx_0|}$

Let $\displaystyle \delta=min \left\{ 1, \frac{\epsilon}{|xx_0|} \right \}$ for any given $\displaystyle \epsilon>0$.

2. You cannot use x in the definition of $\displaystyle \delta_{\epsilon}$ but only $\displaystyle \epsilon$ and $\displaystyle x_0$

3. $\displaystyle |x-x_0|<\delta_{\epsilon} \leq 1$

$\displaystyle x_0-1<x<1+x_0$

Therefore:

$\displaystyle |\frac{1}{x}-\frac{1}{x_0}| < \frac{\delta_{\epsilon}}{|xx_0|}<\frac{\delta_{\ep silon}}{|(x_0-1)x_0|}$

Let $\displaystyle \delta=min \left\{ 1, \frac{\epsilon}{|(x_0-1)x_0|} \right \}$ for any given $\displaystyle \epsilon>0$.

4. First, because this is a new problem you should have started a new thread.

What you have done above has nothing to do with the correct proof.
Conceptually this particular type of problem is one of the most difficult.
As you did note if $\displaystyle s \in \left( {0,1} \right)\; \Rightarrow \;s \ne 0$.
Now we must do some preliminary work. $\displaystyle \frac{s}{2} > 0$ so we can say that
$\displaystyle \left| {x - s} \right| < \frac{s}{2}\; \Rightarrow \; - \frac{s}{2} < x - s < \frac{s} {2}\; \Rightarrow \;\frac{s}{2} < x$.
Because we are working with denominators we have $\displaystyle \frac{1}{x} < \frac{2}{s}$.

Follow this very closely. If $\displaystyle \varepsilon > 0$ then pick $\displaystyle \delta = \min \left\{ {\frac{s}{2},\frac{{\varepsilon s^2 }}{2}} \right\}$ so that
$\displaystyle \left| {x - s} \right| < \delta \, \Rightarrow \,\left| {\frac{1}{x} - \frac{1} {s}} \right| = \frac{{\left| {s - x} \right|}}{{xs}} < \frac{2}{s}\frac{\delta } {s} = \frac{{2s^2 \varepsilon }}{{s^2 2}} = \varepsilon$.