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:


Which simplifies to:


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?

  • $\begingroup$ 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 $8$. $\endgroup$
    – Kenny Lau
    Jun 26, 2016 at 6:17
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    $\begingroup$ 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 $2x$. $\endgroup$ Jun 26, 2016 at 6:18
  • $\begingroup$ 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). $\endgroup$
    – Kenny Lau
    Jun 26, 2016 at 6:18
  • $\begingroup$ 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? $\endgroup$
    – rb612
    Jun 26, 2016 at 6:20
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    $\begingroup$ 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 $2x + h$. $\endgroup$
    – Paramanand Singh
    Jun 26, 2016 at 8:52

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$ Jun 26, 2016 at 11:29
  • $\begingroup$ @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... $\endgroup$
    – Paramanand Singh
    Jun 26, 2016 at 11:36
  • $\begingroup$ @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. $\endgroup$
    – Paramanand Singh
    Jun 26, 2016 at 11:38
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    $\begingroup$ @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. $\endgroup$
    – Paramanand Singh
    Jun 28, 2016 at 5:49
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    $\begingroup$ @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. $\endgroup$
    – Paramanand Singh
    Jun 28, 2016 at 6:00

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


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.

  • $\begingroup$ 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? $\endgroup$
    – rb612
    Jun 26, 2016 at 6:43
  • $\begingroup$ @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. $\endgroup$ Jun 26, 2016 at 7:22
  • $\begingroup$ 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? $\endgroup$
    – rb612
    Jun 26, 2016 at 7:32
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    $\begingroup$ 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. $\endgroup$ Jun 26, 2016 at 7:36
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    $\begingroup$ 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. $\endgroup$ Jun 26, 2016 at 8:14

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