Why limits work

Question

I'm currently a first year student in electrical engineering and computer science. I know how to compute limits, derivatives, integrals with respect to one variable that is things from one variable calculus (mathematics 1). In mathematics 2 we're currently working on series (convergent, divergent, integral criteria, D'Alemberts criteria, Cauchy criteria, absolute convergence ...). English is not my mother tongue, so forgive me I spell something wrong or have grammar mistakes. I'll try to explain my questions as best as I can. I have multiple questions, but they are all intertwined. Since all these things "need" limits, they are my main confusion.

1. I understand the intuition behind the limit and the epsilon-delta definition, but why it works in practice. That is why can I say when computing the derivative of for example $x^2$ is $2x$? In $\lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ I can't just put $0$ since I would get $\frac{0}{0}$, which would be the "true" derivative, because I don't know what that is. After some manipulation I would get $\lim_{\Delta x \rightarrow 0} 2x + \Delta x$ and since $\Delta x$ goes to $0$ that would be equal to $2x$. But this $\Delta x$ will never be $0$, at least as I look at this and from the definition of the limit it would say that I can make $\Delta x$ as close to $0$, but not equal to, if I'm willing to make $x_1$ and $x_2$ as close to each other. Why can I now take this $2x$ and say for instance that the derivative of someone's position it time is $2x$ that is its velocity is $2x$ and not $2x +$ some small $\Delta x$?

2. When trying to see if an infinite series (which never ends) converges or diverges why can I look at a sequence of partial sums (infinite) of that series and based on their convergence or divergence say if the whole series diverges or converges?

3. When I come to professors and ask these and such questions they tell me why am I bothering my self with such question and that I should take it for granted. Then I just want to kill my self. I mean haven't I came here to study how and why things work? I would like it more if they would just tell me that if it is some "higher" or more complex part of mathematics and that I will learn about it later or that it just isn't know why it works the way it works. So should I even continue to study these things, since I will always come across something that I wouldn't be able to understand (since these "basic" limits are confusing me) and all these professors and academia will tell me that I shouldn't worry why it works the way it works and that I should just take it for granted.

4. All the theorems used to proof derivative, integral, convergence, divergence etc. use in one way or another limits. But in the definition of the limit it says that I can make some $f(x)$ as close to some value L, but not equal to it, as long as I'm willing to make $x$ as close to some value $c$. This definition is supposed to be mathematical rigorous, but using these as close don't "look" rigorous to me.

Please help me since I don't know should I even continue with my studies since there is always some mathematical proof which I cannot understand and is preventing me to go forward and that way I'm always lacking behind and everybody expects to understand everything the first time I hear it. I will be grateful for all comments and suggestions.

Answer 1

A function $f:\>x\mapsto f(x)$ given by some expression has a "natural" domain of definition $D(f)$: the set of all $x$ in the realm of discourse (${\mathbb R}$ or ${\mathbb C}$, say) for which $f(x)$ can be evaluated without asking questions. In most cases $f$ is continuous throughout $D(f)$, which means that for all $x_0\in D(f)$, when $x$ is sufficiently near $x_0$ then $f(x)$ is very near to $f(x_0)$.

Now some $f$'s may have "exceptional points" where they are not continuous, e.g., the sign-function, which is defined on all of ${\mathbb R}$, but is discontinuous at $0$. Above all, the set $D(f)$ may have "real" or "virtual" boundary points, where $f$ is a priori undefined. But nevertheless we have the feeling that $f$ has a "reasonable" behavior in the neighborhood of such a point. Examples are $x\mapsto{\sin x\over x}$ at $x=0$ (a "real" boundary point of $D(f)$), or $x\mapsto e^{-x}$ when $x\to\infty$ (here $\infty$ is a "virtual" boundary point of $D(f)$).

All in all the concept of "limit" is a tool to handle such "exceptional", or: "limiting", cases. An all-imortant example is of course the following: When $f$ is defined in a neighborhood of $x_0$ we are interested in the function $$m:\quad x\mapsto{f(x)-f(x_0)\over x-x_0}$$ which has an "exceptional" point at $x_0$. It is impossible to plug in $x:=x_0$ into the definition of $m$.

This brings me to your point 4. which gets to the heart of the matter. I'd rewrite the central sentence as follows: In the definition of the limit of $f(x)$ for $x\to c$ it says that I can make $f(x)$ as close to the value $L$ as I wish, as long as I'm willing to make $x$ sufficiently close to $c$. The idea is: While it is in most cases impossible to put $x:=c$ in the definition of $f$, we want to describe how $f$ behaves when $x$ is very close to $c$.

You then go on to say that "this definition is supposed to be mathematically rigorous, but using these as close and sufficiently close don't look rigorous to me".

The whole $\epsilon$-$\delta$ business serves exactly the purpose to make the colloquial handling of as close and sufficiently close that you are lamenting rigorous.

Life would be simpler if we could define $\lim_{x\to c}f(x)=L$ by the condition $|f(x)-L|\leq |x-c|$, or maybe $|f(x)-L|\leq 100|x-c|$. But four centuries of dealing with limits have taught us that the $\epsilon$-$\delta$ definition of limit, arrived at only around 1870 or so, captures our intuition about them in an optimal way. It takes care as well of the unforeseeable cases when the error $|f(x)-L|$ can be made as small as we we want, but we need an extra effort in the nearness of $x$ to $c$, e.g., $|x-c|<\epsilon^2$ instead of ${\epsilon\over100}$.

Answer 2

For definiteness, let's start with the modern definition of limits. If $f$ is a real-valued function defined on some deleted neighborhood of the real number $c$, then we say "$\lim(f, c) = L$" if:

For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - c| < \delta$, then $|f(x) - L| < \varepsilon$.

Many people find it helpful to view this definition as a set of rules for an adversarial game. A function $f$, a location $c$, and a prospective limit $L$ are given as above. Player $\varepsilon$ issues a "challenge" in the form of a positive real number. To "meet" the challenge is to ensure that $|f(x) - L| < \varepsilon$ for all $x$ lying in some deleted neighborhood of $c$. The opponent, Player $\delta$, consequently attempts to issue a "response": to specify a positive real number $\delta$ such that every location $x \neq c$ with $|x - c| < \delta$ satisfies $|f(x) - L| < \varepsilon$.

To say "$\lim(f, c) = L$" is to say Player $\delta$ has a winning strategy against a perfect opponent; that is, Player $\delta$ can respond to an arbitrary challenge. This is precisely what is meant by saying, "We can make $f(x)$ as close to $L$ as we like by taking $x$ sufficiently close to $c$ (not equal to $c$)."

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Limits are unique: If $f$ and $c$ are given, at most one number $L$ satisfies the preceding definition. Indeed, if $L_1$ and $L_2$ both satisfy the definition, then for every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - c| < \delta$, then $$ |f(x) - L_1| < \varepsilon/2\quad\text{and}\quad |f(x) - L_2| < \varepsilon/2. $$ (This is a standard analytic idiom; pick a $\delta_1 > 0$ that "works" for $L_1$, pick a $\delta_2 > 0$ that "works" for $L_2$, and let $\delta = \min(\delta_1, \delta_2)$.)

Now pick an arbitrary $x$ with $0 < |x - c| < \delta$. By the triangle inequality, $$ |L_1 - L_2| = |L_1 - f(x) + f(x) - L_2| \leq |f(x) - L_1| + |f(x) - L_2| < \varepsilon/2 + \varepsilon/2 = \varepsilon. $$ But this inequality is a statement about two (fixed) real numbers, and if $|L_1 - L_2| < \varepsilon$ for every $\varepsilon > 0$, then $L_1 = L_2$.

Practically speaking, if we show that some limit is $2x$, then that same limit cannot also be $2x + \Delta x$ unless $\Delta x = 0$.

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Here's how the formal definition works for computing the derivative of $g(x) = x^2$: Fix a real number $c$, define $$ f(h) = \frac{g(c + h) - g(c)}{h},\quad h \neq 0, $$ and put $L = 2c$. (The definition only allows us to "test" a prospective limit; to use the definition we must guess the limit in advance. Here we might notice that $f(h) = 2c + h$ for $h \neq 0$; if we wishfully set $h = 0$, we obtain our guess for $L$. As yet we've proven nothing; for all we know, this guess is incorrect.)

Now let's play the formal limit game: Player $\varepsilon$ issues a challenge. Player $\delta$'s goal is to find $\delta > 0$ such that if $h \neq 0$ and $|h| < \delta$, then $$ |f(h) - L| = \left|\frac{g(c + h) - g(c)}{h} - 2c\right| = \left|\frac{(c + h)^2 - c^2 - 2ch}{h}\right| = \left|\frac{h^2}{h}\right| = |h| < \varepsilon. $$ From this "scratch work"/strategizing, Player $\delta$ discovers they can win by responding with the challenge. That is, if $\varepsilon > 0$ is arbitrary, there exists a $\delta > 0$ (specifically, $\delta = \varepsilon$ in this example) such that if $0 < |h| < \delta$, then $$ |f(h) - L| = \cdots = |h| < \varepsilon $$ (because $\varepsilon = \delta$).

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If you're dealing with an infinite series, the role of the function is played by a partial sum of the series, viewed as a function of the index: $$ s(n) = s_n = \sum_{k=1}^n a_k. $$ To say the series has sum $s$ is to say that for every $\varepsilon > 0$, there exists a positive integer $N$ such that if $n \geq N$, then $|s - s_n| < \varepsilon$. This definition is a challenge-response game of exactly the same type as the "real" limit game. The "function value" $f(x)$ becomes $s_n$, the prospective "limit" is $s$ (which, again, must be known in advance), the "challenge" is $\varepsilon > 0$, and the "response" is an $N$; the response "wins" if $n \geq N$ implies $|s - s_n| < \varepsilon$.

(The condition "$n \geq N$" replaces "$0 < |x - c| < \delta$"; very loosely, this condition asserts "$n$ is closer to $\infty$ than $N$ is". The condition $|s - s_n| < \varepsilon$ is the direct analogue of $|f(x) - L| < \varepsilon$.)

In case these remarks are helpful:

1. A value $f(x)$ may or may not be equal to the limit $L$. Similarly, a partial sum $s_n$ of an infinite series may or may not be equal to the sum $s$.

2. You cannot determine whether a series converges or diverges by looking at a bounded number of terms; you must look at arbitrary finite sums. This is analogous to the impossibility of determining a limit of a function by examining only finitely many points of the domain. (Calculus exercises that as you to "evaluate" a limit by plugging small numbers into a calculator are intuitively compelling, but logically without content. Existence and evaluation of $\lim(f, c)$ can never be rigorously determined by looking at values of $f$ at finitely many points.)

3. You've doubtless seen "convergence tests" for infinite series, and you may know that while $$ \sum_{k=1}^\infty \frac{1}{k^2}\quad\text{and}\quad \sum_{k=1}^\infty \frac{1}{k^3} $$ both converge (i.e., have finite sums), the sum of the first is $\pi^2/6$ while the sum of the second is "unknown" (at this writing). But recall, the definition of convergence requires that the limit be known in advance. The loophole is this: Consider the set of real numbers $S = \{s_n\}_{n=1}^\infty$, with $$ s_n = \sum_{k=1}^n \frac{1}{k^3}. $$ It's easy to prove that $S$ is non-empty and bounded above. It follows by the "completeness property" of the real number system that $S$ has a "supremum" or "least upper bound" $s$, the smallest real number that is greater than or equal to every number in $S$. It's also straightforward to prove $s$ is the sum of the series. When we say "$s$ is unknown", we simply mean there is no known formula for $s$ in terms of familiar numbers (such as $e$, $\pi$, or roots of rational numbers).

Answer 3

One way of explaining how $2x+\Delta x$ "becomes" $2x$ is by applying the standard part function. This function, denoted "st", discards the infinitesimal term $\Delta x$, so that we get $\text{st}(2x+\Delta x)= 2x$. From this point of view, the derivative is not the ratio $\frac{\Delta y}{\Delta x}$ but rather the standard part thereof, namely $f'(x)=\text{st}\left(\frac{\Delta y}{\Delta x}\right)$.

One can explain the limit of a sequence $(u_n)$ this way, as well. Here taking the limit involves two steps: (1) evaluating the sequence at an infinite value of the index $n=H$, say, and (2) taking standard part.

For example, the limit of $(\frac{1}{n})$ is the standard part of $\frac{1}{H}$. Since $H$ is infinite, $\frac{1}{H}$ is infinitesimal, and the standard part of each infinitesimal is $0$. Thus $$\lim _{n\to\infty}\frac{1}{n}=\text{st}\left(\frac{1}{H}\right)=0.$$

Edit 1. The OP wrote:

"After some manipulation I would get $\lim_{\Delta x \rightarrow 0} 2x + \Delta x$ and since $\Delta x$ goes to $0$ that would be equal to $2x$. But this $\Delta x$ will never be $0$, at least as I look at this and from the definition of the limit it would say that I can make $\Delta x$ as close to $0$, but not equal to [it]".

The explanation is that the computation of the limit using the real numbers is an indirect procedure. First you need to guess the value of the limit by trial and error or some heuristic procedure, and only then construct a proof, typically an epsilon-delta one, that this number does the trick. Here epsilon-deltas are a roundabout way of accounting for the disappearance of the remainder term $\Delta x$. On the other hand, if one views the limit as a 2-step procedure as described above (first evaluate at an infinite index, and then apply the standard part to discard the remaining infinitesimal), one gets a direct procedure for getting rid of the remainder term which allows one to guess the answer and prove that it is the correct simultaneously. The proof coincides with the heuristic stage.

Answer 4

I have read the answers and most of them are very high quality, but still I think I can contribute a little bit. I hope this is helpful.

1– I suppose you know the rigorous proof that the derivative of $x^2$ (that is, the limit of $((x+h)^2-x^2) / h$ as $h\to 0$) is $2x$.Suppose that the position of a particle in some reference frame is given by $x^2$ when exactly $x$ seconds have passed.

The reason why you can say that its velocity is $2x$ and not $2x+\Delta x$ is a matter of the definition of the word velocity. In theoretical physics the velocity is defined as the limit value just described when $h\to0$ and not as any value of the ratio for any positive value of $h$ whatsoever.

It is a bit paradoxical that you will never be able to really measure this velocity: even if you measure the position at two very very close moments and compute the ratio, you will always be getting that small $\Delta x$ as a perturbation in your calculation. The surprising part is that if you improve your measurement techniques and decrease the interval between measurements you will actually get something closer to $2x$ every time.

The fact that you are certain about this, that the values will get closer and closer to $2x$ when you approximate the ratio by smaller and smaller $\Delta x$ (which is what you prove mathematically with the $\epsilon-\delta$ definition) is what motivates you to define the "real", "instantaneous" velocity to be equal to $2x$. This velocity is not a ratio of any distance actually traveled in any time interval, but a limit value of these ratios.

Why that mathematical, abstract, definition works in practice is a totally different question, and I'm afraid no one has a complete answer for it. After all, it is a product of reason, not of observation. The physicist Eugene Wigner wrote a famous essay on these topics, called The unreasonable effectiveness of mathematics in the natural sciences. Some religious people like to think that god actually created the physical world with mathematical rules and laws which can be discovered by us, others like to accept that it just works in most cases (not quantum or relativistic scenarios) as practice shows and are happy with that, others just avoid the question.

Just a thought: isn't the assertion "the position of the particle at time $x$ is $x^2$" equally abstract? Even if you measure the position a billion times per second, and ignoring the measurement errors, you would be taking a leap of faith by believing that the position at the infinite number of seconds that you didn't measure obey the same rule. What is usually "best possible" in the natural sciences is to be sure of things up to a certain point and then assume this will hold always by inductive reasoning.

2– Again, it's not that you "can do this", it's that what you quote is the definition of an infinite sum. No one can count up to infinity, and no one can sum an infinite number of terms. The symbol $\sum_0^\infty a_n$ is nonesense until we agree to what we mean by it! If you think of that symbol as representing the limit value of the sequence of partial sums when it exists, you can assign a meaning to it: then you have a new toy to play with, and you can learn how to play with it.

What you want when defining something is that this definition captures desirable properties: that you can treat and operate with infinite sums just the way you treat and operate with a normal sum, and that it will work. As an engineer, you will probably also want it to be useful, to help you solve problems like differential equations, for example.

If you prove that this definition shares the same properties as the normal, finite, sum, then you can play with this new toy just the same way as with your old one. To make sure that this is the case, you prove the theorems that say that $\sum_0^\infty(a_n+b_n)=\sum_0^\infty a_n + \sum_0^\infty b_n \,$ , and so on. If you study and understand the proofs of these theorems, you will see where and how the definition is used, and you will be able to appreciate why it was created the way it was.

3– I think intellectual curiosity is a very important thing to have. You don't have to necessarily finish your search with your professors, you will probably find that different people you meet will be able to provide answers for different kinds of questions. In that sense, it is good that you are asking this here. It will never hurt if an engineer knows his mathematics well, has asked himself profound questions and has tried to answer them seriously. I think it will probably make you be a better professional (or a better human being) if you find your own path through learning.

4– This fourth question has been addressed by several people here, I may not have a better answer than many of them. I can only insist that the rigor in the as close as you wish comes in the $\epsilon-\delta$ definition of the limit and the fact that you can prove that the limit is a unique, well defined object.

Answer 5

If I am understanding correctly your question is essentially: How can we say that a limit is $\textbf{equal}$ to something when it really only ever $\textbf{gets very close}$?

This is because of how we have chosen to $\textbf{define}$ a limit. It is a matter of convention you could say. To emphasise this consider what your epsilon delta definition says:

"$\textbf{We say}$ that the limit of $f(x)$ as $x$ goes to $a$ $\textbf{is equal}$ to $L$ if and only if for all $\varepsilon >0$ there exists a $\delta>0$ so that $|f(x)-L|<\varepsilon$ whenever $|x-a|<\delta$."

Now suppose we have the function $f(x)=x$ and I ask you the following question:

What is the unique number $L$ that will let me make $|f(x)-L|$ as small as I like if I am taking $x$ very close to $0$ $\textit{ie.}$ if I take $|x-0|<\delta$?

The number that works in this case is $L=0$. So what does our definition say about this $L$? It says that the limit of $f$ as $x$ goes to $0$ $\textbf{is equal}$ to $L$. Or put in a more compact notational form $\lim\limits_{x \to 0}f(x)=L=0$.

When we say that the limit of a function is equal to something we mean it is equal to the number $L$ with the properties stated in the definition.

Answer 6

Interesting applets:

A Delta-Epsilon Applet

Epsilon-delta definition of the limit

Answer 7

We need questioning people like you in engineering and mathematics and the sciences. Sixty years ago and ever since I have had exactly the same problem.

Here is the idea of a limit that you had not been told. Some functions have a value for some particular value of the variable but it cannot be found by substituting in the value of the variable because the function is not defined for that value. Consider the function x/x. This is not defined at x=0 because any value times 0 is 0. But this function approaches 1 as x->0 from both sides, so we invented the idea of a limit that is a more general way to determine the value of a function at a point. That's the essence of it. It's wonderfully simple and ingenious. Now consider the function 1/x. As x approaches 0 from the right, the functions approaches infinity and from the left it approaches negative infinity; so this function has no value at x=0. But the function 1/(x^2) approaches infinity from both directions so even though infinity is not a number, the function is defined at x=0 in a manner of speaking. The business with delta x is simply an analytic way to get at it. Delta x is an artifice; we are not really concerned with its value.

I hope this helps and that you keep insisting that everything makes sense. We need people like you.

Answer 8

First sorry this is a little long... but,

I think this might help you. It's one of the things we learned in our introduction to analysis course when we were doing limits.

Lemma

If $x \in \mathbb{R}$ such that $\forall \delta \geq 0, \; 0\leq x < \delta$ then $x=0$.

Proof: Suppose otherwise, $x>0$.

Then let $\delta=\frac{x}{2} \implies 0\leq \delta =\frac{x}{2} <x$ which is a contradiction to the property we assumed $x$ has.

$ \therefore x=0$.

Once this is proved we tend to just accept it but it is kind of important to note.

So this will help explain 1. to you a bit. Firstly the limit is just that, a limit. It's not a function of $\Delta x$, though it can be handy to plug in values close to the number you're approaching to get a feel for it. The idea would be that the limit should be independent of the variable used in taking the limit ($\Delta x$), it should "take" the value it's tending to.

In the early definition of continuity, without epsilons and deltas, you say a function of one real variable is continuous at $c \in \mathbb{R}$ if:

$\lim\limits_{x\rightarrow c} f(x)=f(c)$. In this case our limit has no $x$'s, it's just a number (hopefully) if we choose a function continuous at c. Then you can talk about a function being the derivative of another function if they agree at each point in some set/interval.

It's the same for our derivative. One of the nice ways to define a differentiability of a function is to say that a function $f$ is differentiable at $c$ if there exists a continuous function $g$ at $c$ such that:

$g(x)=\begin{cases} \frac{f(x)-f(c)}{x-c} \quad & \text{ if } x\neq c \\ f'(c) \quad & \text{ if } x=c \end{cases}$

So if $f$ is differentiable at $c, f'(c)$ exists. If we didn't define the second case $x=c$ separately in the definition of $g$ and the limit existed (and was finite), then we'd call $c$ a removable singularity of $g$. Basically we know $g$ "has" a value there but the formula doesn't allow us to sub in directly. Kinda what you were talking about with the $\frac{0}{0}$ problem.

I think Apostol mentions this in some of his calculus books or maybe it's the one on Mathematical Analysis.

Then you work on making these things rigorous with epsilons and deltas like other people have posted so I won't repeat (hopefully I haven't already). The way they work is the same as with the lemma, you bound things and show you can get arbitrarily close to some limit and therefore the function assumes the limit at the "point" (which might be infinity, like in infinite sums the limit is to infinity).

I think others have said things about convergence of sums so I'll leave that for now. So there is a lot to this stuff, so don't feel too bad if you're stuck with parts of it, I get stuck quite alot too.

Hoping this is useful and also without typos/errors :)

Answer 9

It seems to me you are confusing the concept of a sequence with the concept of a limit. Suppose $(x_n)_{n\in \mathbb{N}}$ is a convergent sequence of real numbers and $x=\lim \limits_{n\to \infty} x_n$ the limit of this sequence. This limit $x$ is a real number and is something completely different as a sequence (formally, a sequence $(x_n)$ is a map $f:\mathbb{N}\rightarrow \mathbb{R}$ where we denote $f(n)$ by $x_n$). It is also not the same as an element of the sequence.

In your first point you seem to confuse the two. You're computing the derivative of $x^2$ using the limit definition and ask why the derivative is $2x$ and not $2x + \Delta x$. The reason is that the derivative is the limit, so a real number. To be precise, a function $f$ is differentiable in $x_0$ if for all sequences $(h_n)$ such that $\lim \limits_{n\to \infty} h_n=0$ the limit $\lim \limits_{n\to \infty} \frac{f(x_0+h_n)-f(x_0)}{h_n}$ exists and is the same for all these sequences (this is the same as saying $\lim \limits_{h\to 0} \frac{f(x_0+h)-f(x_0)}{h}$ exists; alternatively, you could use an $\epsilon-\delta$ definition). This limit is then called the derivative. If we now compute the derivative of $x^2$ using this language, we get: $$\lim \limits_{n\to \infty} \frac{(x+h_n)^2-x^2}{h_n}=\lim \limits_{n\to \infty}2x+h_n$$

Your question then boils down to: why is the derivative of $2x$ and not $2x +h_n$ for some $n$? Well, because the derivative is the limit of the sequence and this limit is $2x$. It doesn't matter that $h_n$ never becomes 0, it suffices that it gets arbitrarily close to 0, because the limit is something distinct from the sequence and its elements.

So until now I have used notions such as closeness that you have criticized in your 4th question. But as Christian Blatter points out, there are rigorous definitions to make these notions precise, namely the definition of a limit and the $\epsilon-\delta$ definition. So if you prove your theorems using these definitions, there is no problem.

A final remark about infinite series: you ask why you look at the convergence of the sequence of partial sums. Well, that is because an infinite series is defined to converge iff the sequence of partial sums converges. That is, for a sequence $(a_n)$ we define the infinite sum of the sequence as: $$\sum \limits_{n=1}^{\infty}a_n:=\lim \limits_{N\to \infty}\sum \limits_{n=1}^{N}a_n$$ The term infinite sum can be a little misleading, since it is not really a sum, but the limit of a sequence.

Answer 10

Before we think about limits of functions in a calculus sense, maybe it makes sense to talk about limits in a topological sense. For simplicity, in this post assume I'm talking about the reals with the usual metric.

A limit point, in the topological sense, is a point $x$ in some topological space $X$ ($\subset \mathbb{R}$, if we desire) such that every neighborhood of $x$ contains a point in $X$ other than $x$. In other words, no matter how small a ball we draw around $x$, we can find some other point in that ball $y \in X$ such that $y \neq x$.

Now, this definition doesn't give you any information about where the point $y$ might be in relation to $x$ or any other such information that might be extracted from a function. But, what it does do is it gives us a way to construct a sequence. Remember that the definition of a limit point affords us the freedom to choose a neighborhood of any size. So for a decreasing sequence of radii $\epsilon_1 > \epsilon_2 > \cdots$, we can find points $y_1, y_2, \ldots$ that get closer and closer to $x$.

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Now, let's talk about functions. Suppose our set $X$ was the image of a function $f(t)$. And further suppose that $t$ is a limit point of $T$, and hence we can find a sequence of radii $\delta_1 > \delta_2 > \cdots$ that induces a sequence $t_1, t_2, \ldots$ that approaches $t$. If our function has some nice properties (specifically, the properties we define with these $\epsilon-\delta$ definitions!), then we can say that the sequence $t_1, t_2, \ldots$, when transformed by $f$, gives rise to a sequence $f(t_1), f(t_2), \ldots$ that approaches $x$!

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This topological definition is handy because it doesn't care about the symbolic quirks that lead to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ situations. Either the sequences exist, or they do not.

When we say something like $\lim_{t\to a} f(t) = a$, what we're really asserting is the existence of a sequence in the domain that induces a sequence in the range that satisfies the definition of a limit point. In other words, we're declaring that there is a "path" in the range of $f$ that leads to the point $f(a)$. That path may be difficult or impossible to construct, but we're asserting that it is there.

This same notion may also elucidate why the partial-sum definition of infinite series makes sense. Each partial sum evaluates to a number. We construct a sequence of these numbers. If the sequence of these numbers forms a path to a single point as we add more and more numbers to the sequence, then we assert that the infinite series converges.

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In short, the easiest way to think about limits is to momentarily divorce yourself from functional notation and thinking, and instead think about space. Finding limits is like finding a set of instructions of how to get to that point; or more appropriately, proving that such a set of instructions does in fact exist. This notion also formalizes the un-rigorous presentation of "a number very close to blah." We're not just choosing a number that's almost "blah", we're choosing a nested set of punctured neighborhoods that are guaranteed to be non-empty, and we're choosing elements from those neighborhoods. It's just that we don't know what those neighborhoods are most of the time, only that they exist.

Answer 11

What follows is something I posted in the Math Forum discussion group ap-calculus on 13 March 2006 that might be useful.

I recently came across a short note on how to convey the idea of a limit to students that I thought readers of this list would find useful. The full text of this short note is given below.

James Clyde Bradford, Anecdotes for limits, School Science and Mathematics 59 #3 (March 1959), 218.

In trying to give beginning students an intuitive feeling for what happens when a limit such as $\lim_{x \rightarrow 1}\frac{x^2 - 1}{x-1}$ is being evaluated, we often say that one may not let $x$ take the value $1$ since the function is not defined at that point. Further, the value at a single point has nothing to do with the limit. However, the function being defined for near values, we may take $x$ close enough to $1$ to see what the function is doing near $1.$ Two anecdotes illustrate the point. The first came from my calculus teacher Prof. Amos Barksdale, North Texas State, the other came from some preacher.

1. I put on my seven league boots and start to town. These boots are such that each succeeding step carries me half the remaining distance to town. I never get to town but I get close enough to do my shopping.

2. Letting $x$ be equal to $1$ is like committing sin. We won't actually commit sin, but we will get close enough to see what it's like.