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Using the $\epsilon$-$\delta$ definition show that $f(x) = \frac 1 {x^2}$ is a continuous function at any $x_0 = a, a > 0$.

To what I understand of this question, is it just asking to me prove that it is continuous using epsilon and delta? I am a little bit confused because it says at any $x_0 =a$, what is this supposed to mean?

If it is just asking for continuity, can I express it as this?:

$$\lim_{x\to a}\frac1{x^2}=\frac1{a^2}$$

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Since you reposted this question including Mathematician 42 answer I suppose that you might need some extra help.

So far we have $$|f(x) - f(a)| = \Big|\frac{(x-a)(x + a)}{x^2a^2}\Big|.$$ You have three terms to play with on the RHS:

  1. $x + a$ is innocuous.
  2. $x - a$ is important: $x$ is approaching $a$, hence you can make this quantity as small as you want choosing $\delta$ appropriately. This is the term that you should exploit in order to bound everything by $\epsilon$.
  3. $x^2a^2$ is a technical complication: if you allow $x$ to go sufficiently close to $0$ the denominator will blow-up. To solve this issue you need to do something clever with $\delta$: notice that, since $a > 0$, if $\delta \le \frac a2$, then $x \ge \frac a2$, i.e. it is bounded away from $0$.

Now you should have all the ingredients to conclude the proof!

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Let $I$ be an open interval in $\mathbb{R}$. A function $f:I\subset \mathbb{R}\rightarrow \mathbb{R}$ is continuous in $a\in I$ if for all $\varepsilon>0$ there exists a $\delta>0$ such that $|x-a|<\delta\Rightarrow |f(x)-f(a)|<\varepsilon.$

We are presented with the function $f:\mathbb{R}_0\rightarrow \mathbb{R}:x\mapsto \frac{1}{x^2}$. We have to show that $f$ is continuous in any $a\in \mathbb{R}_0$. So take such an $a$. Choose $\varepsilon>0$. We have that $|f(x)-f(a)|=|\frac{1}{x^2}-\frac{1}{a^2}|=|\frac{a^2-x^2}{x^2a^2}|=|\frac{(x-a)(x+a)}{x^2a^2}|$. You have to make this expression smaller than $\varepsilon$ by taking an appropriate $\delta$.

Edit: So first of all let's reduce this problem a bit. You probably know that if $f$ and $g$ are two continuous functions, then $f\cdot g$ is continuous as well. Now obviously $f(x)=\frac{1}{x^2}=(\frac{1}{x})^2$. So we only have to show that the function $f(x)=\frac{1}{x}$ is continuous.

Following the same line of attack we find that we need to make the expression $$|\frac{1}{x}-\frac{1}{a}|=|\frac{x-a}{xa}|$$ smaller than $\varepsilon$ by choosing $\delta$ properly. If $|x-a|< \delta$ then $|\frac{1}{x}-\frac{1}{a}|<\frac{\delta}{|ax|}<\varepsilon$ if and only if $\delta<\varepsilon |ax|$. Now this a problem since $\delta$ should not depend on $x$. So we have to resolve this issue.

Now if some $\delta$ works a smaller one will also work, so we may assume that $\delta<\min\left\{1,\frac{|a|}{2}\right\}$. So in particular, if $|x-a|<\delta$, then $|x-a|<\frac{|a|}{2}$. Now think about this last inequality, can you find some bounds for $|x|$ then? If so, you also find an upperbound for $\frac{1}{|x|}$ and you can solve the problem.

Try to think about this, if you cannot find it, I will give one more step.

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  • $\begingroup$ Well, that's the thing, how to make it smaller, i.e. what the appropriate $\delta$ is. $\endgroup$ – peter.petrov Apr 28 '16 at 13:36
  • $\begingroup$ Okay, this one is a bit tricky, I'll edit my answer and give the OP some new ideas. My answer probably isn't that much of a hint yet. $\endgroup$ – Mathematician 42 Apr 28 '16 at 13:41

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