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I think I've finally been able to voice my confusion when it comes to derivatives and limits.

Let's first look at the difference quotient for a function $f(x)=x^2$

$$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$

This becomes:

$$\frac{(x+h)^2-x^2}{h}$$

Which simplifies to:

$$\frac{h(2x+h)}{h}$$

However, when we cancel the $h$, we get:

$$\lim_{h\to0}2x+h\\ h≠0$$

But then how are we able to say that the slope at a point (let's say $x_0 = 4$) is EXACTLY $8$ rather than saying it's approaching $8$? Saying that it is exactly $8$ means that $h=0$ which violates the domain restriction on $h$.

Therefore, it makes more sense to me to take the nonstandard analysis approach with the derivative being $2x+\epsilon$ where $\epsilon$ is some infinitesimally small quantity. $h$ can be arbitrarily small, yes, because of epsilon-delta, but still can never be zero. So can you please explain why we are able to say that the slope is exact?

rb612
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  • Because it can get arbitrarily close to $8$ when $h$ gets arbitrarily close to $0$, so if it can indeed reach $0$, then the slope would be $8$. – Kenny Lau Jun 26 '16 at 06:17
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    Do you know what a limit is? We can formally assign exact values (rather than approximations) to limits. The idea is that by taking $h$ small enough, we can make the difference quotient arbitrarily close to $2x$. – MathematicsStudent1122 Jun 26 '16 at 06:18
  • It's like the hole in the graph of $y=\dfrac{x^2-1}{x-1}$. You can never directly go to the hole, but by looking around it, you can say that it only makes sense if the hole is (1,2). – Kenny Lau Jun 26 '16 at 06:18
  • Yes, and I do understand how the limit can be exact because it's how it's defined. However, we define the derivative as the limiting value. But because it's a limit and not an exact function value, I don't see how we can say that it's exactly 8 for example, not just arbitrarily close. There is a difference between arbitrarily close and equal from what I understand. Another example, if we said it was exactly $8$, that wouldn't that mean we're defining the tangent line from one point rather than the limit of the secant line? – rb612 Jun 26 '16 at 06:20
  • see http://math.stackexchange.com/a/1834183/72031 – Paramanand Singh Jun 26 '16 at 07:13
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    You need to stop thinking of limits and limiting value in terms of "everyday" meaning of these words. They have a well defined meaning in mathematical contexts. Your statement "limits can be exact" should be replaced by "limits are always exact like $2 + 2 = 4$ and not like $\pi \approx 22/7$". Since derivative is also a limit it is an exact real number if it exists. The expression $2x + h \neq 8$ if $x = 4$ and $h \neq 0$ but $\lim_{h \to 0} 2x + h = 8$ if $x = 4$. The expression $\lim_{h \to 0}2x + h$ is something very different from $2x + h$. – Paramanand Singh Jun 26 '16 at 08:52

2 Answers2

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After reading some comments by OP to the other answer, I think that I need to expand my comment (to OP's question) into a full answer.

The fundamental issue here is that OP thinks $\lim_{x \to a}f(x)$ to be on the same level as $f(a)$. But this is not the right approach. In fact the limit may exist (and often does) even when $f(a)$ is not defined.

OP reasons that $\lim_{h \to 0}(2x + h)$ only approaches $8$ and is not exactly $8$ when $x = 4$. I think a crudely correct statement nearest in meaning to the last sentence is this:

Exression $2x + h$ approaches $8$ and is not exactly $8$ when $x = 4$ and $h$ approaches $0$. On the other hand the expression $\lim_{h \to 0}(2x + h)$ is a number which is dependent on $x$ so that it is $8$ when $x = 4$ but it is something totally independent of $h$.

It is important to understand that $L = \lim_{x \to a}f(x)$ is something totally independent of $x$ and instead dependent on $a$ and $f$ (this part is not that difficult to believe) but at the same time time it is also totally independent of $f(a)$ (this is very hard to accept for beginners, in fact $f(a)$ may or may not be defined). The dependence of $L$ on $a$ and $f$ is related to values of $f$ at points near $a$ and the dependence is defined using the usual technical definition of limit involving $\epsilon, \delta$.

The issue with slope of tangent is that many beginners think that there is a definition of tangent to a curve at a point which uses ideas fundamentally different from the concept of derivative. This is another deep misconception and probably it stems from the fact that before calculus the only curve-tangent stuff is the "tangent to a circle" which has a definition not dependent on derivative. A tangent to a circle at point $P$ can be defined as a line passing through $P$ and perpendicular to the radius passing through $P$ (it can also be defined as a line which intersects the circle at only one point and that point is $P$). For curves other than circle such a definition is not available.

Even in the case of circle the definition based on geometry matches with the definition based on derivative. For a general curve the slope of tangent to a curve is defined by the derivative of an appropriate function and hence by definition the derivative equals the slope of tangent.

user21820
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Paramanand Singh
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  • Two comments: 1. It's misleading to speak of $f(a)$ in your response to OP because the function need not even be defined at $a$, which is precisely the case with the difference quotient, as you know -- but that is what OP is missing. 2. The tangent can easily be defined for all quadratics in two variables (conics), not just circles, using an algebraic technique similar to the case I considered in my answer, a parabola. – symplectomorphic Jun 26 '16 at 11:29
  • @symplectomorphic: agree with your second comment. For the first comment I am looking at the final expression $2x + h$ which is defined at $h = 0$ also. The problem which OP faces is not because the difference quotient is not defined at $0$, but his main argument is that $\lim_{h \to 0}2x + h$ can not be equal to $8$ if $x = 4$ it can only approach $8$ and I have tried to explain flaw in this reasoning. Anyway the idea of limit should not be thought as a way to assign values to functions even at points where it is not defined. contd... – Paramanand Singh Jun 26 '16 at 11:36
  • @symplectomorphic: Limit at a point does not depend on whether the function is or is not defined at that point, but it requires that the function be defined at nearby points and it is a statement about the trend of these values at nearby points. – Paramanand Singh Jun 26 '16 at 11:38
  • the difference quotient is not $2x+h$; it is not defined at $h=0$. – symplectomorphic Jun 26 '16 at 12:10
  • @symplectomorphic: I did not say that difference quotient is $2x + h$. – Paramanand Singh Jun 26 '16 at 12:12
  • (I'm not confused about what a limit means. But your answer is misleading for speaking of $f(a)$, which has nothing to do with the limit and indeed may not even exist. The definition requires finding a $\delta$ so that $0<|x-a|<\delta$ implies $|f(x)-L|<\epsilon$. The requirement that $|x-a|$ be positive tells us we do not care what happens at $a$; the function need not be defined there.) – symplectomorphic Jun 26 '16 at 12:15
  • Well you shouldn't be looking at something that isn't the difference quotient, because the question is about the limit of the difference quotient. – symplectomorphic Jun 26 '16 at 12:16
  • @symplectomorphic: I am really not sure what your point is. I know that you know what limits are (it is evident from your answer which I upvoted). I am saying that most beginners try to think of limit as a concept similar to the value of a function and that is the intent of my statement containing $f(a)$. – Paramanand Singh Jun 26 '16 at 12:21
  • My point is that the OP does not grasp that the limit of a function at a point makes sense even when the function is not defined at that point. You try to say this by saying the limit does not depend on $f(a)$, but by using the symbol $f(a)$ you are inadvertently undermining your point: you are tacitly assuming $a$ is in the domain, when it doesn't have to be (and indeed it isn't in the case of the difference quotient). – symplectomorphic Jun 26 '16 at 12:30
  • @symplectomorphic: OK I got it, i will update my answer and let me know if thats OK. – Paramanand Singh Jun 26 '16 at 12:31
  • @ParamanandSingh thank you very much for your answer. +1. I explained some more in the chat room here: http://chat.stackexchange.com/rooms/info/41700/discussion-between-rb612-and-symplectomorphic and you are both helping me to understand this a ton and I really appreciate that - I'm just not at 100% understanding yet. Thank you. – rb612 Jun 27 '16 at 06:40
  • @ParamanandSingh the idea behind my disconnect in understanding I think I really captured. Say you had the infinite Riemann Sum and the smaller the rectangles get, you get a value of X-0.0000001 (some arbitrary error) for the area. But we say that the Riemann Sum equals X. I think of the "error" of the sum as a function of the size of the rectangles (or at least in a direct relationship), which essentially it is. Therefore, when error is zero, then so is rectangle size, and this is not possible. Same idea with derivative. Am I making sense at all? I really appreciate your help both of you. – rb612 Jun 28 '16 at 05:06
  • To elaborate a tiny bit further, in order to yield the exact area under a curve, the rectangle width would need to be zero, but that's obviously impossible. But if width is infinitesimally small, that's treated as zero also in the reals, so I don't see how we can say semantically that it is exact when it seems like a paradox to me. – rb612 Jun 28 '16 at 05:28
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    @rb612: The statement "in order to get exact area the width of rectangle should be zero" is false simply because when when width is $0$ then area is $0$. Unless you understand this fact you will have problems like these. You are still thinking of limits as a value of the function. It is not the case and you have to drop that line of thinking. – Paramanand Singh Jun 28 '16 at 05:49
  • @rb612: It is better that you ask your question about area / integral and I will answer separately. I don't want to mix the handling of area in this answer which deals with slope and derivative. – Paramanand Singh Jun 28 '16 at 05:52
  • You're right. I see how the limit can differ from the value of the function. I have no problems with limits on their own. It's when we define something that is a limit to be exact. Limits themselves can be exact I see, because of epsilon-delta. However because we can never reach the limit in the cases we are talking about, I don't quite see how we act like we do. Like with a derivative, we can never possibly reach it. But we define it as the value we can never reach. Same with integration. You'll never reach exactness. So that's difficult for me to understand. – rb612 Jun 28 '16 at 05:54
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    @rb612: If you are able to see how limits are different from value of a function then you will not be asking this question. Limit concept is not intuitive and has been developed after centuries of efforts of mathematicians. So you may think you understand it, but from your questions it looks like you don't. For example "why do you want to reach the limit?" If I define something $A$ as a limit of some function $f$ as $ x\to a$ then what is problem if $f$ never reaches $A$. for example $0 = \lim_{n \to \infty}1/n$. It doesn't matter if $1/n$ never reaches $0$, but $0$ is exact. – Paramanand Singh Jun 28 '16 at 06:00
  • I appreciate your attempt to pinpoint the misunderstanding. I see the limit as describing what something is getting closer towards, but not necessarily reaching. But I see a difference between arbitrarily close and equal, which I can tell from what you're saying is not the case. Like a secant line approximating the derivative can get arbitrarily close to the limiting value, but the fact that we define derivative as being that limiting value is mind-boggling to me. I see the limit as describing the behavior of a function as it moves towards a value, not as an base to define other things. – rb612 Jun 28 '16 at 06:07
  • And the issue for me is not that we define derivative as being this, we can define it that way and it's fine. The issue is its use as the slope of the tangent line. To say the slope is exactly a value implies to me that somehow we've reached the point where the limit equals the value of the function, which for the derivative, is impossible. – rb612 Jun 28 '16 at 06:12
  • Let us continue this discussion in chat. – Paramanand Singh Jun 28 '16 at 08:20
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The whole point is that the limit of the difference quotient is the definition of the derivative at a point. In general, there is no independent standard against which to check this definition. As a result, it does not make sense to ask "does the value of the limit really give the slope of the tangent line?" -- precisely because we have no other definition of "slope of the tangent line," in general. Put more informally, there can be no question of whether the limit is "exactly" equal to something else, because we have no definition of that something else.

However, perhaps you do think you have an independent definition of "tangent line." If so, you probably have in mind the intuitive idea that a tangent line ought to intersect the curve "at one point." Of course this is wrong in general (think about tangent lines to $y=x^3$ away from $x=0$), but you can check that the limit definition gives the "right" answer (in this sense) for the simplest possible case of $y=x^2$ by a standard precalculus argument. Suppose the line $y=ax+b$ intersects the parabola $y=x^2$ at exactly one point, $(x_0,y_0)$. Then the equation

$$x^2-ax-b=0$$

ought to have precisely one solution (namely $x_0$). But the solutions are $x=\frac{a\pm\sqrt{a^2+4b}}{2}$. Since we have only one solution, the discriminant vanishes, giving $x_0=\frac{a}{2}$. In other words, $a=2x_0$, so the slope of the tangent line $y=ax+b$ is indeed the same as the limit of the difference quotient.

The problem is that this simple geometric picture won't carry you very far: it fails, as I said, even for the nice curve $y=x^3$. (Things can get a whole lot worse: consider the problem of making sense of the tangent line to the curve $y=x^2\sin(1/x)$ at $x=0$.) So we must take a different approach to defining "slope of the tangent line" for general functions. The approach we take is to say a function has a tangent line if the limit of the secant slopes exists.

Why is this a good approach? The short answer is that it allows us to prove many important theorems. But note, as well, that it does generalize some of our geometric intuition, which says that the secant lines ought to slide continuously through the tangent line as $h$ changes from positive to negative. In other words, the limit definition tells us how to define the function $$g(h):=\begin{cases}\frac{f(x+h)-f(x)}{h}& h\neq0\\??&h=0\end{cases}$$ if we want $g$ to be continuous at zero.

symplectomorphic
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  • Wow! This is a great answer, thank you. So then for the second part of my question about having $2x+\epsilon$, it seems like that idea is generally frowned upon, even though it's how Leibniz defined a tangent line. Why can't it touch the curve at 2 imperceptibly close points, in this case $8$ and $8+\epsilon$? Or at least why is this looked down as the wrong way to think about it? – rb612 Jun 26 '16 at 06:43
  • @rb612: I don't know what you're asking. But I remembered seeing a question like this recently, and lo and behold, you asked it. My answer here is very close in spirit to joriki's there. – symplectomorphic Jun 26 '16 at 07:22
  • Yeah, they're all the same variation of the same question. But I think about it a lot, and I think what bothers me is that the limit is arbitrarily close but never necessarily equal. So therefore there's a small error in value that is negligible, yes, and rounded off to a real number is zero, but can't be zero because our original difference quotient has a hole at zero. So let me propose this: Leibniz defined the tangent line as the line through a pair of infinitely close points on the curve. It seems like in real analysis, this definition is bad. It seems to make sense. Can you explain why? – rb612 Jun 26 '16 at 07:32
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    What exactly do you mean when you say the limit is "never equal"? What two quantities are you saying are not equal? What is it that the limit is not equal to? If you dwell on this question you will see that you do not have a definition of what the limit of the difference quotient is "supposed" to be equal to. That is the whole point of defining the slope of the tangent line to be the limit of the difference quotient. You can't worry about whether the limit is "right" (or "equal" to something else) unless you specify exactly what you want to check the limit against. – symplectomorphic Jun 26 '16 at 07:36
  • hmm, good point. Well I'm saying the limit can be equal to something but we never actually reach the limit value when we're dealing with the difference quotient. It's not possible. But we say we reach it. For example, the limit is equal to 8. No problem. But we say the slope is 8. We define the derivative as the limit. So the slope of the line tangent to the curve is the limit of the difference quotient at that point. But it still is strange we can do this when the limit is different than the function value and say it's exact rather than "the best we can do" – rb612 Jun 26 '16 at 08:03
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    Every statement you make is riddled with errors and imprecisions. You say the limit is "different than the function value" -- but what function are you talking about? It is true that the difference quotient itself is never equal to the limit of the difference quotients, but this is not interesting. (Similarly, the sequence $1/2, 1/3, 1/4, ...$ does not contain zero, but its limit is zero.) It would be correct to say that nearby secant slopes approximate the tangent slope, in the sense that the difference quotient for nearby points approximates the limit of the difference quotient. – symplectomorphic Jun 26 '16 at 08:14
  • But you also seem to be going one step further and worrying not about the difference between the difference quotient and the limit of the difference quotient but between (1) the limit of the difference quotient and (2) the slope of the tangent line. The problem is that you have not given a definition of (2). Indeed that is the whole problem: how do we define (2)? The answer is that we define it to be (1). So there is nothing to worry about: (1) is exactly equal to (not an approximation of) (2) by definition. Your problem is that you think you can make sense of (2) by some other definition. – symplectomorphic Jun 26 '16 at 08:17
  • Let us continue this discussion in chat. – rb612 Jun 27 '16 at 06:30