How to understand the definition of limit?

Question

Definition 1: Let $f(x)$ be defined on an interval that contains $x=a$. Then we say that, $$\lim_{x\to a}f(x) = L$$ if for every number $\epsilon$ there is some number $\lambda$ such that $$|f(x) - L|<\epsilon \text{ whenever } 0<|x-a|<\lambda.$$

I understand what this theorem is trying to tell me in terms of mathematical conditions. I also understand the vertical and horizontal line test. The range mapped by the domain has to be within something called 'the pink area', but I still don't see the big picture of it or I don't see the importance of that so called 'pink area'. What does close enough mean?. I have taken several calculus classes and still don't understand a single thing about it.

What was the original problem that led to this problem?

Answer 1

I have taken several calculus classes and still don't understand a single thing about it.

That's sad. Here's a related kind of example that may help.

Suppose you would like to calculate the square root $x$ of $2$. If all you need is an answer within $1$ of the correct answer, then using any value between $x=1$ and $1.5$ will do, since $1^2 \le 2 \le 1.5^2$. If you need an answer that's within $0.1$ of the true value, you will need a tighter inequality on $x$. If you need two decimal place accuracy you will have to restrict $x$ to an even smaller interval. But you can always manage that. If you specify in advance the accuracy you require (typically, that will be some small number traditionally named "$\epsilon$") then it is possible to restrict the value of $x$ to an interval small enough to guarantee that any $x$ in that interval, when squared, will be a good enough approximation - that is, within $\epsilon$ of the true value. We usually use $\delta$ to specify an amount that $x$ can differ from the true value and still guarantee a square withing $\epsilon$ of $2$. It's pretty clear that the smaller $\epsilon$ you start with the smaller $\delta$ will have to be. But you can always find such a $\delta$. That's what it means for the squaring function to be continuous.

Perhaps think of yourself as a consultant asked by a client for a number whose square is close to $2$. To help him out, you ask first "how close to $2$ must the square be?". Until he gives you a value, you can't give him a number. Once he specifies his tolerance, you go back to your desk and compute, and tell him he can use any number between $a$ and $b$ and he'll be safe. Of course you can't find $a$ and $b$ until you know his requirement.

Answer 2

Your outlined text is not a theorem, but a definition.

As Ethan Bolker has said, it is sad that after so many attempts you "still understand a single thing about it".

The basic notion is continuity. A function $f$ is continuous at a point $a\in{\rm dom}(f)$ if inputting an $x$ sufficiently near $a$ into $f$ results in a value $f(x)\approx f(a)$. Of course this intuitive idea has to be expanded into a clear mathematical definition. A simple idea would be to require that $|f(x)-f(a)|\leq |x-a|$, i.e., that the output error is at most as large as the input error. We would also be content with $|f(x)-f(a)|\leq C\,|x-a|$ with a certain constant $C$, say $C=20$. It has turned out that a such so-called Lipschitz condition is not sufficient to capture, e.g., the continuity of $x\mapsto\sqrt{x}$ $(x\geq0)$ at $a=0$, and so we were led to this $\epsilon/\delta$-business.

Now limits are an exception handling measure. Given a function $f$ there often are points $a\in{\mathbb R}$, or $a\in\{-\infty,\infty\}$, that barely miss belonging to ${\rm dom}(f)$. Drawing the graph of $f$ we get the impression that when $x$ is near $a$ the function behaves "reasonably" and "tends to some limit" $\eta$. Now we want to express this observation in a mathematically precise way which then is satisfyable by a real proof. The natural definition is the following: We say that $\lim_{x\to a}f(x)=\eta$ iff defining $f(a):=\eta$ would make $f$ continuous there (when $a\in\{-\infty,\infty\}$ special speak is necessary). Expanding this idea we are again back at the $\epsilon/\delta$-business.

Consider as an example the function $f(x):={\sin x\over x}$ which by this expression is defined for all $x\ne0$. In possession of the exact notions (and a precise definition of $\sin$) we can prove that $\lim_{x\to0} f(x)=1$, and then the extended function (called sinc) is defined and continuous on all of ${\mathbb R}$.