Formal Def of Limit: Why is |x-c| < δ and not |x-c| <= δ

[Hammad Shahid]

Homework Statement

Taking notes on the formal definition of limits.

Relevant Equations

lim (as x -> c ) = L
δ > 0 ; ϵ > 0

*Given: δ = |all real numbers|
*Given: ϵ = |all real numbers|

For any x value within +/- |δ| of c, we can find a y=f(x) within the corresponding +/-|ϵ| of L. According to my professor, the mathematical representation of this is |x-c| < δ and |f(x) - L | < ϵI fail to understand why it cannot be equal to ϵ or δ.

 

[Hammad Shahid]

Is it because this all applies when x=/=c but gets infinitely closer to it.

 

[fresh_42]

There is no mathematical reason, they both work. The reason why "<" is the common case, is that we deal with open sets here, so the definition goes without the boundary. The most common definition of continuity in topological spaces (which are defined by stating what the open sets are) is that the preimage of any open set is open. And open in $\mathbb{R}^n$ means "<". It is a convention rather than a necessity.

 

[PeroK]

Is it because this all applies when x=/=c

Yes.

 

[member 587159]

The following statements are all equivalent:

$$\forall \epsilon> 0: \exists \delta > 0: \forall x: 0 < |x-c| < \delta \implies |f(x)-L| < \epsilon$$
$$\forall \epsilon> 0: \exists \delta > 0: \forall x: 0 < |x-c| \leq \delta \implies |f(x)-L| < \epsilon$$
$$\forall \epsilon> 0: \exists \delta > 0: \forall x: 0 < |x-c| < \delta \implies |f(x)-L| \leq \epsilon$$
$$\forall \epsilon> 0: \exists \delta > 0: \forall x: 0 < |x-c| \leq \delta \implies |f(x)-L| \leq \epsilon$$

This is actually a very good exercise to see that you understand how quantors work, so you should try to prove hese equivalences.

 

[WWGD]

Notice if the result applies when $|x-c|< \epsilon$ it must also apply in any closed subinterval by definition ,e.g., in $|x-c| \leq \epsilon/2$. It also ties down with the definition of continuity at a point $x$ :for every open 'hood $U_{f(x)}$ of $f(x)$ there is a 'hood $V_x$ of $x$ with $f(V_x)\subset U_{f(x)}$

 

[Stephen Tashi]

According to my professor, the mathematical representation of this is |x-c| < δ and |f(x) - L | < ϵ

The correct definition uses $0 < |x -c| < \delta$ This condition excludes the case $x = c$.

For example, define the function $f(x)$ by:
$f(x) = 2x$ if $x \ne 3$
$f(3)= 47$

The way the definition $\lim_{x \rightarrow 3} f(x)$ is written, we have $\lim_{x \rightarrow 3 } f(x) = 6$.
If we wrote the definition using "$|x - c| < \delta$" , a person could argue as follows.

$lim_{x \rightarrow 3} f(x)$ is not $L = 6$, because for $\epsilon = 0.1$ , the value $x = 3$ does not satisfy $| f(x) - 6| < 0.1$. We have $|3 - 3| = 0 < \delta$ , but $|f(3) - 6 | = |47 - 6| = 41 > 0.1$ no matter how small $\delta > 0$ is chosen.

If this is the first time you have encountered limits of functions, you might prefer that $\lim_{x \rightarrow 3} f(x)$ not be 6 because $f(x)$ is not 6 when $x$ is exactly 3. However, it has proven useful in mathematics to define the $lim_{x \rightarrow c} f(x)$ in such a way that it does not depend on a value at $f(c)$.

For example, using the correct definition, we can consider limits such as $\lim_{x \rightarrow 0} \frac{\sin(x)}{x}$ where the function has no defined value at $x = 0$.

For any x value within +/- |δ| of c, we can find a y=f(x) within the corresponding +/-|ϵ| of L.

The way you expressed the idea ("within"), you didn't exclude the case $x = c$.

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I fail to understand why it cannot be equal to ϵ or δ.

As other replies indicate, using "$\leq$" instead of "$\lt$" wouldn't make any difference.

The reason "$\lt$" is used is matter of culture and tradition. In advanced mathematics, we consider functions that are "mappings" between sets. Introductory Calculus deals with functions such as $f(x)= x^2$ , which are mappings from a real number to another real number. There can be mappings of a more general type. For example the function defined by $F((x,y)) = (2x,xy)$ maps a point $(x,y)$ in 2 dimensional space to another point in two dimensions. In the more general case, we define limits in terms of "open sets" of points. If you are familiar with the notation "(a,b)" to indicate an "open interval" on the real number line, you have some hint of the use of "open sets".

In the general case of mappings, the definition of $lim_{x \rightarrow p} F(x) = L$ can be expressed as:
For each open set $E$ containing the point $L$ , there exists an open set $D$ containing the point $p$ such that $F(D - \{p\}) \subset E$. (The notation $F(D - \{p\})$ represents the set $\{F(x): x \in D, x \notin\{p\}\}$ ).

So, in introductory calculus, it is traditional to phrase the definition of $\lim_{x \rightarrow c} f(x)$ so it can be regarded as a statement involving "open sets". This makes the definition follow the same pattern as the definition for limit of a mapping used in advanced mathematics. To use "open intervals", we must use "$\lt$" instead of "$\leq$".

For example, for $\delta > 0$, $\{x: |x-c| \lt \delta\}$ represents the open interval $(c - \delta, c + \delta)$. By contrast, $\{x: |x-c| \leq \delta\}$ represents the "closed interval" $[c - \delta, c + \delta]$.