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I am trying to understand the mechanics of the proof of why $\dfrac{1}{x^2}$ is continuous from $(0, \infty)$.

We want to show $|f(x_0) - f(x)| <\epsilon$ for sufficiently small $|x-x_0|$.

We see $f(x) - f(x_0) = \dfrac{(x_0-x)(x_0+x)}{x^2\cdot x_0^2}$

Now my book says: If $|x-x_0| <\dfrac{x_0}{2}$ then $\dfrac{x_0}{2} < |x| < \dfrac{3x_0}{2}$ and $(x+x_0) < \dfrac{5x_0}{2}$. With this we can conclude:

$|f(x)-f(x_0)| < \dfrac{10|x-x_0|}{x_0^3}$ (Inequality 1)

So take $\delta = \min\{\dfrac{x_0}{2}, \dfrac{x_0^3\epsilon}{10}\}$ and we are done. This is what I do not understand.

I do not understand the reasoning behind choosing delta other than I see that if $|x-x_0| < \dfrac{x_0}{2}$ then if $\delta= \dfrac{x_0^3\epsilon}{10}$ then Inequality 1 will be $\epsilon$ close. However what if $|x-x_0| \geq \dfrac{x_0}{2}$?

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2 Answers 2

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When you’re trying to show that $f$ is continuous at some $x_0$, you care only about how $f$ behaves at values of $x$ close to (but not equal to) $x_0$. Thus, if $x_0>0$, values of $x$ further away than $\dfrac{x_0}2$ don’t matter: they don’t get arbitrarily close to $x_0$, and as a result we don’t care what happens when $|x-x_0|\ge\dfrac{x_0}2$.

For a possibly simpler parallel, consider a convergent sequence $\langle x_n:n\in\Bbb N\rangle$, say with limit $L$. The convergence of the sequence to $L$ doesn’t depend on the first $100$ terms: you can change them to anything you like, and the sequence will still converge to $L$. You can change the first million terms, or the first billion, and the sequence will still converge to $L$. Convergence depends only on tails of the sequence, so as long as you leave a whole tail unchanged $-$ even if that tail doesn’t start until the $10^{10^{10}}$-th term $-$ the sequence will still converge to $L$.

Similarly, $\lim\limits_{x\to x_0}f(x)$ depends only on values of $f$ near $x_0$; you can change $f$ completely outside of any interval $(x_0-\epsilon,x_0+\epsilon)$, and you won’t change $\lim\limits_{x\to x_0}f(x)$ or $f(x_0)$. If the two were equal before the change, they’ll still be equal after the change, and $f$ will still be continuous at $x_0$.

The reasoning in the proof of the continuity of $f(x)=\dfrac1x$ on $(0,\to)$ relies on this. The first step in getting an open interval around $x_0$ in which $|f(x)-f(x_0)|<\epsilon$ is to look at the interval $\left(x_0-\frac{x_0}2,x_0+\frac{x_0}2\right)$; this isn’t small enough to do the job, but it is small enough to let us calculate a $\delta$ that will do the job.

It’s not the only possible choice. Suppose that we start by considering only values of $x$ in the interval $\left(x_0-\frac{x_0}3,x_0+\frac{x_0}3\right)$, i.e., those such that $|x-x_0|<\frac{x_0}3$. Then $\frac{2x_0}3<x<\frac{4x_0}3$, $x+x_0<\frac{7x_0}3$, and

$$|f(x)-f(x_0)|<\frac{|x-x_0|\cdot\frac{7x_0}3}{\left(\frac{x_0}3\right)^2x_0^2}=\frac{21|x-x_0|}{x_0^3}\;.$$

Thus, if you take $$\delta=\min\left\{\frac{x_0}3,\frac{x_0^3\epsilon}{21}\right\}\;,$$

you ensure that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta$. Since $epsilon>0$ is arbitrary, and this shows how to find a satisfactory $\delta$ no matter what $\epsilon$ might be, this shows that $f$ is continuous at $x_0$. And since the computation of a satisfactory $\delta$ depends only on the fact that $x_0>0$, this in turn shows that $f$ is continuous at every positive $x_0$.

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Could you explain the $\delta = \min$ part further? –  CodeKingPlusPlus Nov 28 '12 at 4:22
    
@CodeKingPlusPlus: The calculation that justifies taking $\delta=\dfrac{x_0^3\epsilon}{21}$ depends on knowing that $|x-x_0|<\frac{x_0}3$; if we don’t already know that $|x-x_0|<\frac{x_0}3$, we can’t be sure that taking $\dfrac{x_0^3\epsilon}{21}$ is a small enough value of $\delta$ to ensure that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta$. By setting $\delta$ equal to the smaller of $\dfrac{x_0}3$ and $\dfrac{x_0^3\epsilon}{21}$, we guarantee that $|x-x_0|<\frac{x_0}3$ whenever $|x-x_0|<\delta$, so that making it smaller than $\dfrac{x_0^3\epsilon}{21}$ does what we want. –  Brian M. Scott Nov 28 '12 at 4:27
    
So it is as straightforward as making it what we want so that we can use what we calculated? –  CodeKingPlusPlus Nov 28 '12 at 4:31
    
@CodeKingPlusPlus: Yes: one half of the $\min\{\dots\}$ ensures that $x$ is close enough to $x_0$ for the other half of the $\min\{\dots\}$ to be small enough to do what we want. –  Brian M. Scott Nov 28 '12 at 4:34

The key is your equation $$f(x) - f(x_0) = \frac{(x_0-x)(x_0+x)}{x^2\cdot x_0^2}.$$ We want the right-hand side to have small absolute value.

This will happen if (i) $|x-x_0|$ is small and (ii) the rest of the expression does not spoil things. So we want to make sure that the $x+x_0$ on top is not too big, and that the $x^2x_0^2$ at the bottom is not too close to $0$.

We will want to make $|x-x_0|\lt x_0/2$, for at least a couple of reasons. If $x$ is within $x_0/2$ of $x_0$, then $x$ must be positive and less than $(3/2)x_0$, and therefore $x+x_0$ must be positive and less than $(5/2)x_0$. Also, $x$ must be $\gt x_0/2$. So the $x^2$ at the bottom is greater than $x_0^2/4$. It follows that as long as $x$ is within $x_0/2$, we have $$|f(x)-f(x_0)|\le |x-x_0| frac{10}{x_0^3}.\tag{$1$}$$ (We have done some routine algebra.)

Now suppose that we are given an $\epsilon \gt 0$, and want to make $|f(x)-f(x_0)|\lt \epsilon$. If $|x-x_0| \lt \epsilon\frac{x_0^3}{10}$, then by $(1)$ the desired inequality will hold.

But we have to make sure that Inequality $(1)$ holds. That was obtained under the assumption $|x-x_0|\lt x_0/2$. If by bad luck $\epsilon\frac{x_0^3}{10}$ turns out to be bigger than $x_0/2$, Inequality $(1)$ will not necessarily hold.

But if we choose $\delta$ to be the smaller of $x_0/2$ and $\epsilon\dfrac{x_0^3}{10}$, then everything is OK, we get $|f(x)-f(x_0)|\lt \epsilon$.

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I have a question, why don't we always choose $\delta = \frac{x_0}{2}$ then the inequality will always holds isn't it? –  Mathematics Nov 28 '12 at 4:57
    
@Mathematics: The $\delta$ almost always depends strongly on $\epsilon$. If I want to guarantee that $|f(x)-f(x_0)|\lt 10^{-8}$, almost always I have to choose $\delta$ far smaller than if I just want to guarantee that $|f(x)-f(x_0)|\lt 10^{-2}$. –  André Nicolas Nov 28 '12 at 5:15
    
so taking $\delta=\min\{\frac{x_0}{2},\epsilon\frac{x_0^3}{2}\}$ is to ensure the (1) always holds right? –  Mathematics Nov 28 '12 at 5:26

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