If limit to 0 is never = 0 then how are derivative are precise not approximation?

Question

In all books I've read they initially introduces $dx$ as $\lim_{x \to 0}$ which by definition means that $x \approx 0$ but $x \ne0$.

And derivative as change in $y$ for $dx$ change in x.

But then later they just state that following is precisely equal to slope tangent line at that point. However from the definition of limits it should always give secant line not tangent.

$$ \frac{d}{dx}f(x) = \lim_{\Delta x \to 0} \frac {f(x+\Delta x)-f(x)}{\Delta x} $$

And isn't every line tangent for a point? So what defines the tangent slope we get from the derivation.

Answer 1

Consider the parabola $y=x^2$.

$\Delta y = (x+\Delta x)^2-x^2=2x\Delta x+\Delta x^2$

$\frac{\Delta y}{\Delta x}=2x+\Delta x$

This gives you the slope of the secant line between $(x,y)$ and $(x+\Delta x, y+\Delta y)$

So far this is all algebra, no calculus.

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In keeping with only using algebra, suppose we want to find the slope of the tangent line at the point $(c,c^2)$.

What's the definition of a tangent line? For a circle that's easy, it's a line that only intersects the circle at a single point. This needs to be more general for an arbitrary curve, but for a parabola, we can still use this. The criterion for generalization requires convexity, but is probably a distraction at this time.

Consider the line $y=mx+b$. It has to intersect the parabola at $(c,c^2)$, so

first, where does $y=mx+b$ intersect $y=x^2$?

$x^2=mx+b\implies x^2-mx-b=0 \implies x= \frac{m \pm \sqrt{m^2+4b}}{2} $

The discriminant has to be 0, i.e. $m^2+4b=0$, in order for there only to be one common point between the line and the parabola.

This in turn means $x=c \implies m=2c$. So by definition of a tangent line, we need the discriminant to be $0$. This in turn requires $m=2x$, so the slope of the tangent line to $y=x^2$ is $m=2x$ at point $(x,x^2)$.

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How do you reconcile this with the $\Delta x, \Delta y$ terms above?

$\frac{\Delta y}{\Delta x}=2x+\Delta x$ makes sense so long as $\Delta x \ne 0$. What are we to make of $\Delta x \to 0$? $

A standard, rigorous definition is $\forall \delta>0, -\delta \le \Delta x \le \delta$. Equivalently, the absolutely value of $\Delta x$ is less than any positive value $\delta$.

Using algebra, we've determined $\frac{\Delta y}{\Delta x}-2x=\Delta x^2$

So $-\delta \le \Delta x \le \delta \implies -\delta^2 \le \frac{\Delta y}{\Delta x}-2x \le \delta ^2$

Here, we've gotten rid of an explicit reliance on $\Delta x$ needing to be $0$.

We have proven algebraically that however small $\delta$ is, The difference between the slope of the secant line to $2x$ is even smaller since $0\le \delta \le 1 \implies 0\le \delta ^2 \le \delta \le 1$.

So we do have an approximation, true. Consider though, every approximation has an error term. Suppose we want a secant line which includes the point $(x,x^2)$ which differs from $2x$ by less than $\epsilon >0$.

We know this difference is $\delta^2<\delta x$. So we are guaranteed an error term $<\epsilon$ if we compose our secant line with points close enough to $x$.

What is close enough? $\Delta x \le \epsilon$ suffices for, algebraically that means $-\epsilon^2 \le \frac{\Delta y}{\Delta x} -2x \le \epsilon^2$

And we know $0\le \epsilon^2 \le \epsilon.$ In other words if our approximation to $x$ for the other point on the secant line is close to $x$, the slope of the tangent line is approximately $2x$ with an even smaller margin of error. And since we argue the difference between $x$ and the putative other point on the secant line is smaller than any positive number, it follows that the error of the slope vs. $2x$ is smaller than any positive number.

From here we assert that given an approximation to a quantity, we assert the approximation is equal to the quantity it self if the absolute value of the error is smaller than any positive number.

We justify this regarding the limits of functions because we do this essentially with any irrational number.

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We know $\sqrt{2}$ is irrational. So if we restrict ourselves only to integers or rational numbers, $\sqrt{2}$ does not exist. Let $ A = \{ x \in \mathbb Q, x^2<2\}$

We know $\sqrt{2} \notin A$. We also have that $1 \le x \le \frac{x+2/x}{2}\le \sqrt{2}$

$1\in A$. For any element in $x \in A$, there exists another element greater than $x$ also in $A$. It follows that $A$ has no highest element even though it is bounded. That means $\forall \epsilon >0, \exists x\in A, \sqrt{2}-\epsilon<x<\sqrt{2}$

Thus we can find an element in $A$ that approximates $\sqrt{2}$ with arbitrary accuracy, i.e. given any positive quantity, we can find an element of $A$ which approximates $\sqrt{2}$ more accurately than that. While $\sqrt{2} \notin A$, for any positive error margin, we have an element in $A$ close enough.

This applies to any irrational real number. Any irrational number can be approximated arbitrarily closely by some subset of rational numbers. So we can think of the set of rational numbers to have holes, quantities greater than or less than but never equal to any rational errors, yet so close to rational numbers that approximating these quantities with "close enough" reals allows you to do arithmetic with the reals. In other words, arithmetic with approximations to real numbers can be used in place of arithmetic of actual real numbers with arbitrary accuracy. This is essentially a rigorous definition of a limit which the earlier arguments generalize to functions over the reals.

So despite being approximations, the approximations are arbitrarily accurate, so we have as much justification asserting their equality as we do the existence of $\sqrt{2}$ or other irrational numbers and for much the same reasons.

Answer 2

It is true what you say that if you have a graph consisting of a single point then it does not make sense to talk about a tangent line. On the other hand, if your number system contains infinitesimals, you can consider the line through a pair of infinitely close points on the curve (meaning that the distance between the points is infinitesimal).

This is still not the tangent line but a secant line, but the tangent line can be obtained from it by taking the standard part, as the following example illustrates. Here the standard part rounds off any finite number to an infinitely close standard number. Consider the parabola $y=x^2$ at the origin. I think everybody would agree that the expected tangent line is the horizontal line $y=0$ (the $x$-axis).

Now consider the following two points on the parabola: $(0,0)$ and $(m, m^2)$ where $m$ is infinitesimal. It is easy to check that the line through the two points has equation $y=mx$ (a line of infinitesimal slope, which is "almost horizontal"). For all finite $x$, the product $mx$ is infinitesimal, and therefore its standard part is $0$. Therefore by applying the standard part to the equation, we get the equation $y=0$ of the tangent line, as expected.

To summarize, the tangent line is not a secant line, but rather the standard part thereof, in the precise sense outlined above.

Answer 3

This is an excellent question. There are other posts on this site that address the issue but none seems to me to match the OPs question, so I will try another.

I think trying to understand motion is a better motivator than trying to understand tangents to curves. That was in fact why Newton invented calculus.

To measure a velocity you measure how far you travel in a particular length of time and divide. So if it takes you an hour to travel $30$ miles you're velocity is $30$ miles per hour. But you don't need to travel for a whole hour to measure that. If you covered $5$ miles in $12$ minutes you could calculate your velocity as $30$ miles per hour (after you correctly converted the minutes to hours). You would get the same result if you traveled $1$ mile in $2$ minutes.

Each of those measurements is just your average velocity over the time interval you used. This is all very well if you are traveling at a constant speed - but you might have been speeding up and slowing down during the time. The calculation wouldn't tell you that.

You (and Newton) might want to think about how fast you were traveling at some particular instant of time. Then the rule "divide the distance covered by the time" would be $0/0$, which makes no sense either mathematically or physically. To make that measurement you could track your average velocity over smaller and smaller time intervals and hope the results clustered around some particular value, which you would then use as the definition of your velocity at that instant.

This strategy turns out to work as long as you assume that you can't "go from $0$ to $60$ instantaneously" - that there is some kind of smooth transition in the way your speed varies.

Now back to the question as asked, about tangent lines. Draw the graph that plots the distance you have traveled from some fixed point as a function of the time it took. Then the slope of the secant line between two points on the horizontal axis is the average velocity over that time interval. The closer together the points, the shorter the interval. If the measured slopes for intervals starting at time $t$ cluster near some number, that is the number we have decided to call the instantaneous velocity at time $t$. In the picture, the secant lines for short intervals cluster around a single line that we define as the tangent line.

When Newton first did this many of his contemporaries asked the same question you posted here. It took mathematicians several centuries to provide a really good answer. That answer is really too complicated to go into when you are first learning calculus. It's good to be aware of the question as you go about learning to use calculus, but not good to focus on it. Just as, when you learn to drive, you don't need to know how the motor under the hood works.