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Let's take my question as an example. I just don't get it.

What does $\delta = \min \{ 2, \frac{\epsilon}{2} \} $ mean ? (especially 'min{}')

Thomas Produit
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user1511417
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  • When you want to proove limits of a function via $\epsilon$ or $\delta$ it really becomes confusing. – user1511417 Jan 04 '15 at 21:46
  • When proving continuity, you are usually served an $\epsilon>0$ and should come up with a $\delta>0$ such that a certain condition (that you do not write) is fulfilled. In this case you (or someone else) claims that it suffices to take $\delta$ as the minimum of the quantities $2$ and $\epsilon/2$. – mickep Jan 04 '15 at 21:48
  • I does take some practice, are you trying to prove that the funciton is continuous everywhere, or just at a particular point? – Tim Raczkowski Jan 04 '15 at 21:48
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    The $min$ function returns the smallest of the input values. For example $min{1,2,3} = 1$ and $min{1,x}$ will return a value of $1$ whenever $x\in[1,\infty)$ and will return a value of $x$ otherwise. For your example, with $\epsilon\geq 4$ you never need to take a $\delta>2$ since $\delta=2$ is sufficient. – JMoravitz Jan 04 '15 at 21:51
  • Related: http://math.stackexchange.com/questions/418961/epsilon-delta-proof-that-lim-limits-x-to-1-frac1x-1/418991#418991 – Clarinetist Jan 04 '15 at 22:05

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It turns out $\delta=\varepsilon/2$ is plenty in this particular case, so you don't need this sort of argument here.

A better example is to show that $f(x)=x^2$ is continuous at $1$. We do some algebra:

$$|f(1)-f(y)|=|1-y^2|=|1+y||1-y|.$$

Since this has a factor of $|1-y|$, the naive thing to do is to make $\delta$ depend linearly on $\varepsilon$. This has a problem: when $\varepsilon$ is large, $\delta$ will be too large, because $|1+y|$ will grow. For example, taking $\delta=\varepsilon/3$ will not work for $\varepsilon=300$, because for $y=101$ we get $|1-y^2|$ being something like $10000$.

So we need to keep $|1+y|$ moderately small, which we can do only by controlling $|1-y|$. One way is to require $|1-y|<1$ regardless of $\varepsilon$. This works because $|1-y|<1$ is equivalent to $0<y<2$, so we guarantee $1<1+y<3$ and in particular $|1+y|<3$.

This means that if we require $|1-y|<1$ then we have $|1-y^2|<3|1-y|$. We want this to be less than our given $\varepsilon$, so we further require $|1-y|<\varepsilon/3$ to get $|1-y^2|<3 \varepsilon/3 = \varepsilon$ as desired.

To get $|1-y|<1$ and $|1-y|<\varepsilon/3$, we take $\delta=\min \{ 1,\varepsilon/3 \}$.

Ian
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  • Wait.. (I'm not used to inequations.) How did you get from $|1-y| < 1$ to $|y|<2$ ? And how can you (all of a sudden) infer from $|1+y| < 3$ that $|1-y| < \frac{\epsilon}{3}$ ? – user1511417 Jan 05 '15 at 10:26
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    For the first one you just solve the inequality: $|1-y|<1$ implies $-1 < 1-y < 1$, which implies $-2 < -y < 0$, which implies $0<y<2$. The second one is not an inference but an additional requirement. The trick is that if $|1+y|<3$ and $|1-y|<\varepsilon/3$ then $|1-y^2|<\varepsilon$. (The $/3$ came about to balance the multiplication by $3$.) We can just require $|1-y|<\varepsilon/3$, while we can get $|1+y|<3$ by requiring $|1-y|<1$. – Ian Jan 05 '15 at 17:30
  • That's more clear! Thank you. I guess you used the third binomial formula here? ($|1-y²|$) Maybe you can add the things in your comment to your answer, to make it easier to understand for other readers/students. – user1511417 Jan 05 '15 at 18:49
  • Solutions like this are still hard for me to detect. I guess you need practice for that. You just don't deal with inequations in secondary school. – user1511417 Jan 05 '15 at 18:53
  • @user1511417 I edited a bit, is this clearer? – Ian Jan 05 '15 at 19:14
  • Yes. Thank you, Ian. It's stil an awkward topic, though. But you can not help it. – user1511417 Jan 06 '15 at 15:54
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What does $\delta = \min \{ 2, \frac{\epsilon}{2} \} $ mean ? (especially 'min{}')

$$\min\left\{ 2, \frac{\epsilon}{2}\right\}=\left\{\begin{array}{ccc}2&\text{if}&\epsilon\gt4\\\frac{\epsilon}{2}&\text{if}&\epsilon\leqslant4\end{array}\right.$$

Did
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Let $\epsilon>0$ be given. Choose $δ:=\min\{\frac\epsilon2,2\}$. Then:

$$\left|f(x)-f(x_0)\right|=\left|\frac{x}{1+x^2}-\frac{x_0}{1+{x_0}^2}\right|=$$

$$=\left|\frac{(xx_0-1)(x_0-x)}{(1+x^2)(1+{x_0}^2)}\right|<\delta\left|\frac{xx_0-1}{(1+x^2)(1+{x_0}^2)}\right|$$

Now let's show that for any $\delta: \space \left|\frac{xx_0-1}{(1+x^2)(1+{x_0}^2)}\right|<2$

( I'll show only $\frac{xx_0-1}{(1+x^2)(1+{x_0}^2)}<2$, it's not hard to show that it's $>-2$ )

$$\frac{xx_0-1}{(1+x^2)(1+{x_0}^2)}<\frac{(x_0+\delta)x_0-1}{(1+(x_0-\delta)^2)(1+{x_0}^2)}$$

Let's assume:

$$\frac{(x_0+\delta)x_0-1}{(1+(x_0-\delta)^2)(1+{x_0}^2)}>2$$

Then:

$$\frac{\delta^2(2x_0^2+2)-\delta(4x_0^3+5x_0)+(2x_0^4+3x_0^2+3)}{(1+(x_0-\delta)^2)(1+{x_0}^2)}<0$$

The denominator and numerator are always positive. Contradiction.

So $\left|\frac{xx_0-1}{(1+x^2)(1+{x_0}^2)}\right|<2$. Then:

$$\left|f(x)-f(x_0)\right|<\delta\left|\frac{xx_0-1}{(1+x^2)(1+{x_0}^2)}\right|<2\delta\leq\epsilon$$

Then if for $\epsilon>2$ we pick $\delta = 2$ and $\delta = \frac\epsilon2$ otherways the inequality holds.

Vladimir Fomenko
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  • That's weird for me. Where do you get the $\delta$ in the second line of your equation? – user1511417 Jan 05 '15 at 10:10
  • And after you discovered the contradiction, you start over again like in the beginning. ("Now let's show...") – user1511417 Jan 05 '15 at 10:11
  • I use the definition of continuity.

    $f$ is continuous in $x_0$ if for any $\epsilon>0$ there is such $\delta$ that for each $x\in D(f), |x-x_0|<\delta \rightarrow |f(x)-f(x_0)|<\epsilon$. So $|x_0-x|=|x-x_0|<\delta$. My proof is based on this definition. I start with $|f(x)-f(x_0)|$, then prove an auxiliary lemma and return to the start.

     – Vladimir Fomenko Jan 05 '15 at 15:56
  • You can read more here: https://en.wikipedia.org/wiki/Continuous_function#Weierstrass_definition_.28epsilon-delta.29_of_continuous_functions – Vladimir Fomenko Jan 05 '15 at 16:03