2
$\begingroup$

Let's consider that we need to prove: if $f(x) = x^2 + x-3$ we have to show that $f(x)\to -1$ as $x \to 1$.

To prove this, we set $\delta = \mbox{min} \{1,\frac{\epsilon}{4}\}$ and argue that $|f(x)-L|<\epsilon$

My question is why do we use a $\mbox{min} \{1,\frac{\epsilon}{4}\}$ rather than a $\mbox{max} \{1,\frac{\epsilon}{4}\}$? How does taking a min satisfy the restriction that we initially make on $\delta$ to bound $|x-1|$?

$\endgroup$

5 Answers 5

3
$\begingroup$

When you limit $\delta$ in more than one way by introducing different upper bounds, you need that all upper bounds are satisfied to ensure that all the steps based on these upper bounds remain valid.

By taking the smallest upper bound, the other ones are automatically satisfied (but this is not the case when taking the largest upper bound!). So if you need: $$\delta \le a_1 \;\mbox{and}\; \delta \le a_2 \;\mbox{and}\; \ldots \;\mbox{and}\; \delta \le a_n$$ then all these inequalities are satisfied by taking: $$\delta \le \mbox{min}\left\{ a_1,a_2,\ldots,a_n \right\}$$

Simple example: if somewhere along the proof you require $\delta \le 2$ and a bit later you also require $\delta \le 1$, then you are sure both inequalities are satisfied by taking $\delta \le 1$, i.e. $\delta \le \mbox{min}\left\{ 1,2 \right\}$. Note that it would not be sufficient to take $\delta \le 2 = \mbox{max}\left\{ 1,2 \right\}$ because then $\delta = 1.5$ would be possible, but that doens't satisfy $\delta \le 1$.

Of course in this example, it is clear that $1 < 2$ so that we could simply take $\delta \le 1$. However, you don't always know which upper bound is the smallest one as it may contain variables such as $\varepsilon$. By using min, we are sure to take the smallest upper bound and hence satisfying all conditions on $\delta$. In your example, the bound will depend on the value of $\varepsilon$; for example:

  • if $\varepsilon = 8$, then $\delta = \text{min} \{1,\frac{\epsilon}{4}\} = \text{min} \{1,\frac{8}{4}\} = \text{min} \{\color{red}{1},\color{blue}{2}\} = \color{red}{1}$
  • if $\varepsilon = 2$, then $\delta = \text{min} \{1,\frac{\epsilon}{4}\} = \text{min} \{1,\frac{2}{4}\}= \text{min} \{\color{blue}{1},\color{red}{\tfrac{1}{2}}\} = \color{red}{\tfrac{1}{2}}$

The smallest upper bound is automatically 'selected', whatever the value of $\varepsilon$.


Remark: note that this is not only the case for limits of quadratic functions (as mentioned in your question). For any limit proof, or more generally even for any context where you need multiple upper bounds to be simultaneously satisfied, this is a way to do it.

$\endgroup$
0
$\begingroup$

You cannot use $max$ in this context because ${\epsilon}$ can be chosen arbitrarily small - so small that eventually $max${1,${\epsilon}/4$} is simply 1. Then you are saying |$f(x)$ - L| < ${\epsilon}$ FOR ALL $x$ in (0,2), regardless of the value of ${\epsilon}$ - which is plainly nonsense : for all $x$ in (0,2), $f(x)$ takes on all values in (-3,3).

$\endgroup$
0
$\begingroup$

Maybe if I fill the heuristic process you shall understand. Set $$ f(x) :=x^2 +x -3,\quad L :=-1. $$ Write backwards: $$ |f(x)-L| <\epsilon \\ \mathrm{...if}\quad |x^2+x-2| =|(x-1)(x+2)| =|x-1 || x-1+3 | <\epsilon. \\ \mathrm{...if}\quad |x-1 || x-1+3 | \leq |x-1 | \cdot ( |x-1|+3 ) \leq \delta \cdot (1+3) =\delta \cdot 4 <\epsilon \\ \mathrm{...if}\quad |x-1| \leq 1 \quad \land \quad |x-1| \leq \epsilon/4. $$ How to make sure both is true? take $\delta$ to be the smaller of them ($1$ and $\epsilon/4$) and assume $|x-1| \leq \delta$.

$\endgroup$
0
$\begingroup$

Because $x<\min\{a,b\}$ implies both $x<a$ and $x<b$, whereas $x<\max\{a,b\}$ implies only that either $x<a$ or $x<b$ (but not necessarily both).

Knowing $x<\min\{a,b\}$ therefore lets you use anything that is a consequence of either of the conditions that implies.

Example: My income from performing a certain task depends on how long I take. If I complete the task within one week, I will receive \$100. But if I do it within one day, I will receive an additional bonus of \$50. Now if I select a completion time of max{1 day, 1 week} = 1 week, I will only get \$100 since the basic payment condition was satisfied but the bonus condition was not. On the other hand, if I select a completion time of min{1 day, 1 week} = 1 day, I will have satified both the basic payment condition and the bonus condition, so I will receive both the payment and the bonus, totaling \$150.

$\endgroup$
0
$\begingroup$

Simply speaking, it's because the $\min$ is more restrictive. If you have property A holding for $|x-a| < \delta_1$ and property B holding for $|x-a| < \delta_2,$ then both properties A and B will hold, simultaneously, for $|x-a| < \min (\delta_1,\delta_2).$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .