How can we say the derivative is exact if the difference quotient has a domain restriction?

Question

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?

Answer 1

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.

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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.

Answer 2

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.