Understanding of the formal and intuitive definition of a limit

Question

The intuitive definition for $\lim\limits_{x \to a} f(x) = L$ is the value of $f ( x )$ can be made arbitrarily close to $L$ by making $x$ sufficiently close, but not equal to, $a$ . I can easily understand this ,but for the (ε, δ)-definition of limit:For every real ε > 0, there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ implies | f(x) − L | < ε. Oh, god, I cannot understand it completely.

As it is the formal definition of limit, I think it should be precise but somewhat should include the mean of the above intuitive definition, so as for "$f ( x )$ can be made arbitrarily close to $L$" in the intuitive definition correspond to “For every real ε > 0,| f(x) − L | < ε” in the formal definition, it's fine! but does “making $x$ sufficiently close, but not equal to, $a$ .” correspond to “there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ” ? This is the point I cannot understand, because I am not sure if “there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ” shows "x close enough, but not equal, to $a$".

The other question is: does the biggest δ also get smaller as ε is getting smaller? Why? (Exclude the case when f(x) is a constant function.)

Why do we need the formal definition of a limit? Does the intuitive definition have some flaw?

P.S. Thank you everyone, but I must declare I only have some basic knowledge of limit, I only started to learn calculus a few days ago.

Answer 1

The rationale behind the concept of limit is exception handling: We have a function $f:\>\Omega\to {\mathbb R}$ defined on some set $\Omega$, and we are given a "place" $a$ which belongs to $\Omega$, or at least is "adherent" to $\Omega$. Therefore the function $f$ may or may not be defined at $a$. But we observe (e.g., by letting Mathematica draw the graph of $f$) that "when $x$ is near $a$ then $f(x)$ is near a particular value $\eta$". If that is the case we'd like to tell this to other people by writing $\lim_{x\to a} f(x)=\eta$.

Now we need a formal definition for such a fact. Under what circumstances would a value $\eta$ qualify as limit of $f(x)$ when $x\to a$? The answer is simple: If defining $$f(a):=\eta\tag{1}$$ (resp. overriding the given definition of $f(a)$ by $(1)$) would make $f$ continuous at $a$.

Now appeal to the definition of continuity: A function is continuous at $a$, if, given any tolerance $\epsilon>0$ we can guarantee $|f(x)-f(a)|<\epsilon$ by choosing $|x-a|$ "sufficiently small", i.e., smaller than a certain allowance $\delta>0$, which will depend on the given tolerance $\epsilon$.

Answer 2

First, notice that

$f(x)$ can be made arbitrarily close to $L$ by choosing $x$ close enough to $a$

is equivalent to

No matter how close you want $f(x)$ and $L$ to be, you can limit the distance between $x$ and $a$ such that $f(x)$ and $L$ will be as close as you wanted.

Let's give names ($\varepsilon$ and $\delta$) to the two "close" distances:

For any maximum distance $\varepsilon > 0$ between $f(x)$ and $L$, there is a maximum distance $\delta > 0$ between $x$ and $a$, such that whenever $x$ and $a$ are closer than $\delta$, $f(x)$ and $L$ are closer than $\varepsilon$.

Removing redundant words, we get:

For any $\varepsilon > 0$, there is a $\delta > 0$, such that whenever $x$ and $a$ are closer than $\delta$, $f(x)$ and $L$ are closer than $\varepsilon$.

And using $|a - b|$ instead of "distance between $a$ and $b$", we get:

For any $\varepsilon > 0$, there is a $\delta > 0$, such that whenever $|x - a| < \delta$, $|f(x) - L| < \varepsilon$.

Answer 3

The $\varepsilon$-$\delta$-definition of the limit can be read like this:
for any neighbourhood of $L$, no matter how small (points with $|y-L|<\varepsilon$), there is a neighbourhood of $a$ (the points with $0<|x-a|<\delta$), such that $f$ maps the entire neighbourhood into the neighbourhood of $L$ (that is $|f(x)-L|<\varepsilon$).
This captures the idea of the intuitive definition. The intuitive way of seeing this makes $\delta$ smaller as $\varepsilon$ gets smaller, since being closer to $L$ is a stronger condition and is satisfied by fewer points around $a$. This is just not required by the definition.
If there is a $\delta>0$ that satisfies $|f(x)-L|<\varepsilon$ for all $0<|x-a|<\delta$ then this also true for all $0<\delta'<\delta$, you can always choose a different $\delta$ that is smaller than all others you have seen before, and imagine $\delta$ being monotonically decreasing. The important point is that $\delta$ is non-zero, that there always is a neighbourhood of $a$ such that all points of it get mapped arbitrarily close to $L$ by $f$.

Answer 4

As you get closer and closer to the x you are approaching, does the function get closer and closer to the limit? Notice there's two things getting smaller here.

In other words, x is arbitrarily close to a, is f(x) arbitrarily close to L?

The point is we have to choose the $\delta$, such that f(x) is as close to L as we need.

Here's an example,

$$\lim_{x \to 2} x^2 $$ and $$ |f(x) - 4| = |x^2 - 4| = |x-2||x+2|. $$

The first thing you should notice is we have a $|x + 2|$ term that us unwanted, so we need to bound it under $\epsilon$. Because we have $0 < |x - 2| < \delta$, this term will take itself out if $\delta < 1$.

Choosing $\delta < 1$, we have $ |x+2| < \delta+4 < \delta + 4\delta = 5\delta $ because $ |x - 2| < \delta $, so we can choose $\delta < min(1,\frac{\epsilon}{5})$ then,

$$ |f(x) - 4| = |x-2||x+2| < |x + 2| < 5\delta \leq 5\frac{\epsilon}{5} = \epsilon.$$

See how I compare $\delta$ to $\epsilon$? I already had $\delta$ in terms of $\epsilon$, so it was trivial to ensure the limit was smaller than $\epsilon$.