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I am studying continuity and in the textbook I follow is given an example with function $x^2$ to be continuous at $x_0=3$. This is how it is explained in the textbook:

Some arithmetic converts this to $$|x^2-9|= |x − 3| |x + 3| \,\,.$$ If we insist that $|x − 3| < \frac{\varepsilon}{M}$, where M is bigger than any value of $|x + 3|$, then we will have

$$|x^2-9| < \varepsilon $$

exactly as we need. But just how big might $|x + 3|$ be? If we remember that we are interested only in values of $x$ close to $3$ (not huge values of $x$), then this is not too big. For example, if $x$ stays inside $(2, 4)$, then $|x + 3| < 7$. These are enough computations to allow us to write up a proof. Let $\varepsilon > 0$. Let $\delta = \frac{\varepsilon}{7}$ or $\delta = 1$, whichever is smaller (i.e., $\delta$ = min{$\frac{\varepsilon}{7}, 1$}). Then if $|x − 3| < \delta$ it follows that $|x + 3| = |x − 3 + 6| ≤ |x − 3| + 6 < 7$ and hence that

$$|x^2-9| = |x − 3| |x + 3| < 7 |x − 3| < 7(\varepsilon/7) = \varepsilon.$$

The thing I don't understand is: why is the assumption made that $x$ stays in the interval $(2,4)$? What if $x$ does not stay in this interval? Will the choice of $\delta$ as $\min\{\frac{\epsilon}{7},1\}$ still work?

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    Yes I edited a few minutes ago. You are right that's the book I'm following. Sorry. – 2017-01-23

3 Answers 3

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The answer is no. If the interval is different, e.g., $x \in (1,5)$, we know, $|x+3| < 8$, hence you'll need $\delta = \min \{\epsilon / 8, 1\}$.

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    So what is the choose of $\delta$ to prove this continuity. It depends on what? – 2017-01-23
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    It comes from the definition of continuity: given an $\epsilon > 0$, there exists a $\delta$ (can depend on both the $\epsilon$ and the point $x_0$ where you want to show the continuity) s.t. for every $x$ with $|x - x_0| < \delta$, must hold $|f(x) - f(x_0)| < \epsilon$. – 2017-01-23
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    ok, ok but for this particular function $f(x) = x^2$ how should I choose $\delta$? – 2017-01-23
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    There is no fixed rule. You have to see what works -- in fact, here you saw that two different choices (and many other choices will) worked. So, it is up to your choice -- both will be given credit -- and makes analysis interesting. – 2017-01-23
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    Ok so that's what I thought too. There's up to my choice. So let's say if I want to study the continuity of $ f(x) = \frac{1}{x} $, at point $x_0=\frac{1}{2}$I can rewrite $|\frac{1}{x}-2|$ as $|\frac{2}{x}| |x-\frac{1}{2}|$. So if I pick $x \in (1,2)$, then $\delta$ should be min$\{1,\frac{\epsilon}{3}\}$ – 2017-01-23
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    $x = 1/2$ does not fall in $(1,2)$. pick an interval that covers it. – 2017-01-23
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    also mark the question as answered. – 2017-01-23
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    Yeah, I think I got it. Thanks – 2017-01-23
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There is a discontinuity at x = 0. So the first thing we need to do is limit $\delta$ such that when $|x-3|<\delta, x>0.$ It is slightly arbirtrary to say assume $\delta \le 1$ but $1$ is easy to work with.

We could have just as easily said assume $\delta \le \frac 12$ or $\delta \le 2$ but had we said let $\delta \le 3$ we would not be keeping clear of that discontinuity and getting ourselves into trouble.

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If $x$ is not in the interval $(2,4)$, then with $\delta=\min\{\frac\epsilon 7,1\}$, we will not have $|x-3|<\delta$, hence need not prove anything about $|x^2-9|$.