Why Does the Epsilon-Delta Definition Not Require |f(x) - L| to Be Nonzero?

[PFuser1232]

I'm trying to wrap my head around the epsilon-delta definition.
"Let $f$ be a function defined on an interval that contains $a$, except possibly at $a$. We say that:
$$\lim_{x →a} f(x) = L$$
If for every number $\epsilon > 0$ there is some number $\delta > 0$ such that:
$|f(x) - L| < \epsilon$ whenever $0 < |x - a| < \delta$"
Why aren't we restricting $|f(x) - L|$ to be nonzero?

 

[pasmith]

I'm trying to wrap my head around the epsilon-delta definition.
"Let $f$ be a function defined on an interval that contains $a$, except possibly at $a$. We say that:
$$\lim_{x →a} f(x) = L$$
If for every number $\epsilon > 0$ there is some number $\delta > 0$ such that:
$|f(x) - L| < \epsilon$ whenever $0 < |x - a| < \delta$"
Why aren't we restricting $|f(x) - L|$ to be nonzero?

|x - a| must be bounded away from zero because the concept of a limit deals with what happens to f near to, but not at, a. It might happen that f(x) = L for values of x near to, but not at, a; this is not a problem.

For example, consider f: (-1, 0) \cup (0,1) \to \mathbb{R} : x \mapsto 1. Then \lim_{x \to 0} f(x) = 1 since for any \epsilon &gt; 0, if 0 &lt; |x| &lt; 1 then |f(x) - 1| = |1 - 1| = 0 &lt; \epsilon.

 

[Stephen Tashi]

Why aren't we restricting $|f(x) - L|$ to be nonzero?

For example, if f(x) is the constant function f(x) = 2 whose graph is a horizontal line, we want \lim_{x \rightarrow 3 } f(x) to exist and be equal to 2.

 

[Svein]

Why aren't we restricting |f(x)−L||f(x) - L| to be nonzero?

Why should we? If |f(x)-L| = 0, it is obviously less than ε.

 

[PFuser1232]

Why should we? If |f(x)-L| = 0, it is obviously less than ε.

So while $|x - a|$ must be nonzero, $|f(x) - L|$ could be zero?

 

[PeroK]

So while $|x - a|$ must be nonzero, $|f(x) - L|$ could be zero?

The point is that the function value at a does not affect the limit. The limit relates to points close to a, but not a itself.

But there's no reason the function value can't be equal to the limit.

 

[PFuser1232]

Take for example $\lim_{x→1} 2x + 3 = 5$.
We want to prove that for every $\epsilon > 0$ there is some $\delta > 0$ such that:
$|(2x + 3) - 5| < \epsilon$ whenever $0 < |x - 1| < \delta$
This seems to be the general approach taken in most texts/websites:
$|2x - 2| = 2|x - 1| < \epsilon$ implying that $|x - 1| < \frac{\epsilon}{2}$ which "looks a lot like" $|x - 1| < \delta$, so we set $\delta$ equal to $\frac{\epsilon}{2}$.
Next, we plug in $\delta = \frac{\epsilon}{2}$ into the inequality $0 < |x - 1| < \delta$ and multiply throughout by 2. We end up with $0 < |2x - 1| < \epsilon$ which, again, "looks a lot like" the first inequality, which confirms the fact that our guess was right.
Isn't that last step redundant? Haven't we already established that $\epsilon = 2 \delta$ by manipulating the first inequality? Isn't this just doing the same exact thing, except this time it's the other way around?
Also, while the inequalities look the same, there's still a slight difference. We end up with $0 < |f(x) - L| < \epsilon$. Doesn't this contradict $|f(x) - L| < \epsilon$ (to some extent) since the latter places no restriction on $|f(x) - L|$ being zero?
And finally, is there another way to look at it other than "this inequality looks an awful lot like the other inequality, therefore $\delta = \frac{\epsilon}{2}$"? I can't see why $\delta$ and $\frac{\epsilon}{2}$ can't be different despite both being larger than $|x - 1|$.

 

[PeroK]

Also, while the inequalities look the same, there's still a slight difference. We end up with 0<|f(x)−L|<ϵ0 < |f(x) - L| < \epsilon. Doesn't this contradict |f(x)−L|<ϵ|f(x) - L| < \epsilon (to some extent) since the latter places no restriction on |f(x)−L||f(x) - L| being zero?

I just want to address this one point. I ask you to find a number $x$, such that $|x| < 1$. You suggest $x = 0.5$. I then say "no". That $x$ satisfies:

$0 < |x| < 1$ and, therefore, doesn't satisfy $|x| < 1$

Note that:

$0 < |f(x) - L| < \epsilon \ \ \Rightarrow \ \ |f(x) - L| < \epsilon$

So, there is no contradiction there.

 

[PeroK]

This might help. Rewrite the definition of limit (equivalently) as follows:

$\forall \ \epsilon > 0, \ \exists \ \delta > 0$ such that $|x-a| < \delta$ and $x \ne a \ \Rightarrow \ |f(x) - L| < \epsilon$

Does that clarify things at all?

All we are doing in both cases is excluding the point $a$ from consideration.

 

[Svein]

Take for example limx→12x+3=5\lim_{x→1} 2x + 3 = 5.

No. We do not take the limit of an equation. We take the limit of a function. So, if you want an example, take \lim_{x\rightarrow 1}2x+3. First, choose ε: Choose ε=0.1. Then either start calculating or make a guess: Let δ=0.01 and see what happens. On one side: 2*1.001 + 3 = 5.002. On the other side: 2*0.999 + 3 = 4.998. From these two results, we assume that the limit is 5, since |5.002-5| = 0.002 which is less than ε and |5-4.998| = 0.002 which is also less than ε. Now you can either create a formula for δ or you can choose a series of ε's and continue guessing δ's until you have a good idea of how to choose them.

Now, this may seem obvious, but let me give you another example: Let y = floor(x) (the integer part of x). This function is continuous at all points except when x is an integer (and there it is "continuous from above").

 

[HallsofIvy]

I just want to address this one point. I ask you to find a number $x$, such that $|x| < 1$. You suggest $x = 0.5$. I then say "no". That $x$ satisfies:
$0 < |x| < 1$ and, therefore, doesn't satisfy $|x| < 1$

Surely this is not what you meant to say!? 1/2 satisfies both $0< |x|< 1$ and $|x|< 1$.
Indeed, any number that satisfies $0< |x|< 1$ must also satisfy $|x|< 1$ because $0< |x|< 1$ requires that x satisfy both $0< |x|$ and $|x|< 1$.

Note that:
$0 < |f(x) - L| < \epsilon \ \ \Rightarrow \ \ |f(x) - L| < \epsilon$

So, there is no contradiction there.