1
$\begingroup$

Proving that $\frac{1}{x^2}$ is continuous on (0,1)

$\forall \epsilon >0$ , $ \exists \delta>0 $ such that $\lvert x-x_0 \rvert < \delta $ and $\lvert f(x)-f(x_0) \rvert < \epsilon$

So can I take $\lvert\frac{1}{x^2}-\frac{1}{x_0^2} \rvert < \epsilon$

$\lvert\frac{ x_0^2 -x^2}{x_0^2 x^2 } \rvert=\lvert \frac{(x_0-x)(x_0+x)}{x_0^2 x^2} \rvert = \lvert x-x_0 \rvert \lvert \frac{x+x_0}{x_0^2 x^2}\rvert < \delta* \frac{x+x_0}{x_0^2 x^2}< \epsilon$

I'm confused as to what to do now. Am I looking to create my $\delta$ function or do I have to examine individual cases?

$\endgroup$
  • $\begingroup$ You want to find a bound on $|(x + x_0)/x_0^2x^2|$, call it $M$. Then given $\epsilon > 0$, choose $\delta = \epsilon/M$. $\endgroup$ – Simon S Nov 3 '15 at 21:10
  • $\begingroup$ @SimonS what do you mean by "find a bound", set $\lvert \frac{x+x_0}{x_0^2 x^2} \rvert <M$ and use the upper bound? $\endgroup$ – Ryan T. Donnelly Nov 3 '15 at 21:14
1
$\begingroup$

To give an upper bound to

$$\frac{x+x_0}{x_0^2 x^2}$$

first we have $x_0, x\in (0,1)$ so

$$\frac{x+x_0}{x_0^2 x^2}\le \frac{2}{x_0^2 x^2}. $$

But we still want to give a bound to $x$ (we do not need to care about $x_0$ as it is fixed). So we need a lower bound for $x$ (thus an upper bound for $1/x^2$). Then we make an extra assumption that $x > x_0/2$ (or $|x-x_0| <x_0/2$). Then

$$\frac{x+x_0}{x_0^2 x^2} \le \frac{8}{x_0^4}\Rightarrow \left| \frac{1}{x^2} - \frac{1}{x_0^2}\right| \le \frac{8}{x_0^4} |x-x_0|$$

Now it is clear how we choose $\delta$. For all $\epsilon >0$, choose $$\delta = \min\left\{ \frac{x_0^4}{8} \epsilon, \frac{x_0}{2}\right\}.$$

Then whenever $|x-x_0|<\delta$, we have

$$\left| \frac{1}{x^2} - \frac{1}{x_0^2}\right| \le \frac{8}{x_0^4} |x-x_0| < \epsilon.$$

(In the first inequality we used $\delta <x_0/2$, in the second we used $\delta < \frac{x_0^4}{8} \epsilon$). As $\epsilon >0$ is arbitrary, we have $$\lim_{x\to x_0} \frac{1}{x^2} = \frac{1}{x_0^2}$$

and so $\frac{1}{x^2}$ is continuous on $(0,1)$.

$\endgroup$
  • $\begingroup$ why the $\frac{x_0}{2}$ in $\lvert x-x_0 \rvert < \frac{x_0}{2}$ ? Because we are discussing two points? $\endgroup$ – Ryan T. Donnelly Nov 3 '15 at 21:33
  • $\begingroup$ It doesn't really matter. You set it. All you need is an lower bound on $x$. You could, for example, let $x > x_0/3$. So you need $|x- x_0| < \frac{2x_0}{3}$ and you get instead $\frac{2}{x_0^2x^2} \le \frac{18}{x_0^4}$. Then you need to choose a different $\delta$. @RyanT.Donnelly $\endgroup$ – user99914 Nov 3 '15 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.