Why does the Limit Process give an Exact Slope?

[NoahsArk]

TL;DR Summary

Why does the limit process give an exact slope of the tangent line instead of just a very close approximation?

I am taking a summer calculus class now. For years I've been stuck on the question of why the limit process gives us an exact slope of the tangent line instead of just a very close approximation. I don't need to know the reason for this class I'm taking- we are basically just learning rules of derivates and integrals and applying them. However, it bothers me that I still don't understand the reason after this long.

When you take the limit of the rise over run of any function as delta x (which I'll call h) gets closer to 0, you can never make h equal to zero because then you'd have zero in the denominator. Also, if h were zero, you'd have a point and not a line, and a point can't have a slope. Since h can never be zero, how can you ever know the exact slope?

The only explanations I can think of aren't fully satisfying to me, and I'm not sure if they are good ones. One example of an explanation is in taking the slope of $f(x) = x^2$ at the point where x = 2. Using the limit process you would get the slope function $\frac {4h + h^2} {h}$ You can't set h = to zero here but you can get an equivalent function of 4 + h. So the slope at x = 2 is 4. These two slope functions are equivalent but one has a gap and the other one doesn't, so we can say that the slope must be where the gap is since that's the only place where h is zero. This seems a bit hand wavy to me since the simplified slope function of 4 + h is not the exact same function as $\frac {4h + h^2} {h}$.

Is the explanation more trivial then I'm thinking, and is there just some conceptual misunderstanding I'm having. Or, it it actually complicated and the subject of another class like real analysis, in which case it's ok for me not to have more than an intuitive feeling for why the limit process gives us an exact slope? Thanks

 

[PeroK]

The limit isn't a recursive process. It's a single number that has a certain, well-defined property.

 

[NoahsArk]

Thank you for the response @PeroK . I know that the slope is a single number. My question is how do we know for sure what the exact number is. We can never set $\Delta x$ to zero to know an exact number for sure.

 

[fresh_42]

Thank you for the response @PeroK . I know that the slope is a single number. My question is how do we know for sure what the exact number is. We can never set $\Delta x$ to zero to know an exact number for sure.

The slope of the curve at a point is defined as the slope of the tangent at this point, which can be visualized as the limit of the slope of secants defined through the tangent point $p$ and another one $p+\Delta x$ that is approaching it.

$$
f'(p)=\left. \dfrac{d}{dx}\right|_{x=p}f(x)=\lim_{\Delta x \to 0}\dfrac{f(p+\Delta x)-f(p)}{\Delta x}
$$

 

[NoahsArk]

@fresh_42 yes, my confusion is not in what a tangent line is- I'm confused about how we know the value of the slope of the tangent line from using the limit process. I am confused because the process is setting $\Delta x$ to be zero, but you can't have a zero in the denominator. So, while we can get really really close to the slope of the tangent line by setting $\Delta x$ as close to zero as we want, we can never set it to zero. So, how can we ever have anything more than a really close approximation? Thanks

 

[PeroK]

@fresh_42 yes, my confusion is not in what a tangent line is- I'm confused about how we know the value of the slope of the tangent line from using the limit process. I am confused because the process is setting $\Delta x$ to be zero, but you can't have a zero in the denominator. So, while we can get really really close to the slope of the tangent line by setting $\Delta x$ as close to zero as we want, we can never set it to zero. So, how can we ever have anything more than a really close approximation? Thanks

You don't get a limit by setting $\Delta x=0$.

 

[NoahsArk]

@ PeroK, I meant you you get the slope of the tangent line by taking the limit as $\Delta x$ approaches zero of the slope function $\frac {f(x +h) - f(x)} {h}$ (where h = $\Delta x$). My confusion lies in the fact that although very small changes in x will give closer and closer approximations to the slope of a given point, you can't actually set the change in x to be zero without getting a zero in the denominator. My description of what I am getting confused on was poorly worded I think, but I hope I've clarified.

 

[PeroK]

@ PeroK, I meant you you get the slope of the tangent line by taking the limit as $\Delta x$ approaches zero of the slope function $\frac {f(x +h) - f(x)} {h}$ (where h = $\Delta x$). My confusion lies in the fact that although very small changes in x will give closer and closer approximates to the slope of a given point, you can't actually set the change in x to be zero without getting a zero in the denominator. My description of what I am getting confused on was poorly worded I think, but I hope I've clarified.

That's correct. You need to look at the definition of the limit.

 

[Ibix]

Perhaps consider a concrete example. Say $f(x)=x^2$. What is $\frac{f(x+\Delta x)-f(x)}{\Delta x}$ and is there a $\Delta x$ in the denominator?

 

[NoahsArk]

@Ibix I know the answer to this is 2x both from doing the limit process and also from learning the power rule. I also have an intuitive visual sense of why this is the case. However, what I don't know is why it is EXACTLY 2x and not just very close to 2x. Seems like a Xeno's paradox type issue.

 

[Ibix]

@Ibix I know the answer to this is 2x

That's not the answer to the question I asked. The correct answer is that $\frac{f(x+\Delta x)-f(x)}{\Delta x}=2x+\Delta x$. So there's no $\Delta x$ in the denominator and no problem letting it tend to zero.

 

[fresh_42]

@fresh_42 yes, my confusion is not in what a tangent line is- I'm confused about how we know the value of the slope of the tangent line from using the limit process. I am confused because the process is setting $\Delta x$ to be zero, but you can't have a zero in the denominator.

It is not a process, and we are not setting anything to zero. The "process" with the secants was only meant to visualize what a limit is.
$$
\lim_{\Delta x \to 0} \dfrac{f(p+\Delta x)-f(p)}{\Delta x} = s
$$
stands for a logical statement: For any given $\varepsilon >0$ there is a $\delta(p,\varepsilon )> 0$ such that $\left|\dfrac{f(p+\Delta x)-f(p)}{\Delta x} - s\right| <\varepsilon$ whenever $|\Delta x| < \delta(p,\varepsilon ).$

It means that we can get with our quotient - $\dfrac{f(p+\Delta x)-f(p)}{\Delta x}$ - the slope of the secants, as close as we want - $\varepsilon$ - to the slope of the tangent - $s$ - if we are picking our second point - $p+\Delta x$ - close enough - less than $\delta(p,\varepsilon )$ - to $p$.

I wrote $\delta(p,\varepsilon)$ instead of the usual $\delta$ to note that the choice of $\delta$ depends on the choice of $\varepsilon$ and the location of the tangent $p.$

Anyway, this logical statement is neither a process nor did we anywhere replace $\Delta x$ by zero.

Let's see what my picture says:
\begin{align*}
f(x)&=\dfrac{1}{5}x^2 \wedge p=-5\\[6pt]
f'(-5)&=\lim_{\Delta x \to 0}\dfrac{f(-5+\Delta x)-f(-5)}{\Delta x}=\lim_{\Delta x \to 0}\dfrac{1}{5}\cdot \dfrac{25-10\Delta x+(\Delta x)^2 - 25}{\Delta x}=2-\dfrac{1}{5}\lim_{\Delta x \to 0} \Delta x
\end{align*}

Now, we are at a crucial point. Why is $\lim_{A \to 0}A=0$? The logical statement says: We can get with $A$ as close as we want to $0$ if we only are sufficiently close to zero. That is certainly true, and the logical statement can easily be verified. But it is also rather trivial. We have $\varepsilon =\delta$ and it happens to be, that we get the same result if we simply identify $A=0.$ It's a matter of convenience to shorten the verification of the logical statement. And as it works, we can often make that substitution, but only in the last line of my little calculation. We did not divide anything that was zero.

So, while we can get really really close to the slope of the tangent line by setting $\Delta x$ as close to zero as we want, we can never set it to zero.

Correct.

So, how can we ever have anything more than a really close approximation? Thanks

Because we ended up with $\lim_{A \to 0}A$ which equals $0$ because it is a number $L=0$:
$$
\forall \;\varepsilon >0 \; \exists \;\delta:=\varepsilon >0\;:\; |A-L|=|A-0|=A<\varepsilon \;\forall \; |A|< \delta
$$
This is simply a true statement, so $\lim_{A \to 0}A=0.$ We have thus proven that we can compute $L$ in $\lim_{A \to 0}A=L$ by replacing $A$ by $0.$ That makes computations a lot easier. But is one thing to do it here, and another to divide by zero.

 

[NoahsArk]

@Ibix Ok thanks for clarifying. While I am aware that $\frac {f(x + \Delta x)} {\Delta x} = 2x + \Delta x$, this explanation has still been not totally convincing for me because the two slope functions aren't really exactly the same. One of them, $\frac {f(x + \Delta x)} {\Delta x}$ has a gap in the graph where x = 0, and the other one $2x + \Delta x$, has no gap. The first of these though is the real slope function since it is in the form of rise over run. So isn't that the form what we need to use? Why can we get away with creating a new (although very similar) function?

 

[hutchphd]

Is it easier to justify in your head if you take the limit in a more symmetric rfashion? (The result will be the same in the limit.)
-
$$ \frac {f(x+\Delta x)-f(x-\Delta x))} {2\Delta x}$$

Do you not believe that there exists a unique value for the slope??? How else would one find its value?

 

[Mark44]

(Relative to the example $f(x) = x^2$)

However, what I don't know is why it is EXACTLY 2x and not just very close to 2x. Seems like a Xeno's paradox type issue.

Nothing to do with Zeno's Paradox.

This difference quotient below gives the slope of the secant line between the points $(x, f(x))$ and $(x + h, f(x + h))$.
$$\frac{f(x + h) - f(x)}h = \frac{(x + h)^2 - x^2}h = \frac{2xh + h^2}h = 2x + h$$
For any two points x and x + h, the slope of the secant line will be 2x + h.
The derivative of f is the limit of the above difference quotient; that is,
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}h = \lim_{h \to 0} 2x + h = 2x$$

What you seem to be having a problem with is the fact that we have h in the denominator and are taking the limit as h approaches 0. Note that the numerator is also approaching 0 as well. That makes the fraction one of several indeterminate forms -- expressions that may or may not represent a number. Every time you calculate a derivative using the definition of the derivative you're working with an indeterminate form.

The situation here is similar to the situation when we compare the two functions $f(x) = x + 1$ and $g(x) = \frac{x^2 - 1}{x - 1}$. These two functions are almost exactly the same, differing only at exactly one point; namely, (1, 2). If we take the limit of f as x approaches 1, we trivially get 2 for the result. Although the second function, g, isn't defined at x = 1, we can take the limit of g(x) as x approaches 1, and also get a result of 2. The graphs of these two functions are nearly identical, but we say that the latter function has a "hole" at (1, 2).

 

[vela]

@Ibix Ok thanks for clarifying. While I am aware that $\frac {f(x + \Delta x)} {\Delta x} = 2x + \Delta x$, this explanation has still been not totally convincing for me because the two slope functions aren't really exactly the same. One of them, $\frac {f(x + \Delta x)} {\Delta x}$ has a gap in the graph where x = 0, and the other one $2x + \Delta x$, has no gap. The first of these though is the real slope function since it is in the form of rise over run. So isn't that the form what we need to use? Why can we get away with creating a new (although very similar) function?

It's because when you take a limit, you don't care what the functions are doing at $x$ (or if they're even defined there); you only care about what they're doing in the neighborhood of that point. In that neighborhood, the two functions are exactly the same, so they approach the same limiting value as $\Delta x \to 0$.

 

[PeroK]

@Ibix I know the answer to this is 2x both from doing the limit process and also from learning the power rule. I also have an intuitive visual sense of why this is the case. However, what I don't know is why it is EXACTLY 2x and not just very close to 2x. Seems like a Xeno's paradox type issue.

It's important to dissociate the function values from the limit. Believing that the limit must be one of the function values seems to be a stumbling block for many students.

As an example, consider the set $\{\frac 1 n\}$ for natural numbers $n$. Clearly no member of that set is zero. But, zero is a lower bound for that set. And, in fact, it's the greatest lower bound.

We see that zero has an important relationship with that set despite not being a member of the set. This is actually the starting point for the formal study of limits. In fact, we can also say:
$$\lim_{n \to \infty} \frac 1 n = 0$$And, again, zero (precisely zero) is the limit of that sequence without being an element of the sequence.

Limits and derivatives of real-valued functions are developed from this.

 

[mcastillo356]

Definition of limit

Before formally define the notion of limit, we must take a glance at the intuitive meaning. The definition shows two concepts that need more precision.

1. The first one is the meaning of the sentence "$f(x)$ is as close to the limit $l$ as you want".
We say this sentence when, for any positive quantity, the gap between $f(x)$
and $l$ can be taken, in absolute value, less than the mentioned positive quantity.
2. The sentence "$x$ enough nearby to $a$"is the second aspect of the definition that needs to be explained.
Means that, in absolute value, the difference between $x$ and $a$ is less than a positive small quantity.

Note that, in the formal statement, the condition $0<|x-a|$ is mandatory, just to exclude from the definition the point $a$.

I've taken the quote from a maths textbook for Spanish 17 to 18 years old students.
I've got the sensation @NoahsArk that, I am posting to you though you already know what is all about. Anyhow, best wishes!

 

[jbriggs444]

Why does the limit process give an exact slope of the tangent line instead of just a very close approximation?

How do you define the "tangent line"?

In mathematics, notions like a "tangent line" are meaningless until we define them. Having defined them, the meaning is in the definition. There is nothing else.

Once you define the "tangent line" and the "slope" of the tangent line at a point you have, in principle, a number for the slope of the tangent line at a point.

Once you define a "first derivative" of a function as a number (if it exists) that satisfies some "limit" definition at a point you also have a number.

It is likely that you designed both definitions to yield unambiguous results. It is likely that you chose at least one of these definitions so that it would match the other.

It should then be no surprise that the two numbers, so defined, are unambiguous and matching.

 

[DaveE]

Others here have mentioned this, but I'll single out the idea of the epsilon-delta proof and a key concept of calculus, that things that are "arbitrarily close" are essentially equal.

 

[WWGD]

I don't think you're getting an exact

Perhaps consider a concrete example. Say $f(x)=x^2$. What is $\frac{f(x+\Delta x)-f(x)}{\Delta x}$ and is there a $\Delta x$ in the denominator?

There may be an exception , both in the definition, result, when we consider the linear change of a line; linear object ( n-dimensional linear object). Here, the approximation to the change is _ equal_ to the change, as , given say a line$f(x)=ax$ ( ignore the constant; just a translation) the limit quotient becomes $\frac {ax+a(\Delta x)-(ax)}{\Delta x}=\frac{a\Delta x}{\Delta x}$ which equals $a$ as long as $\Delta x \neq 0$. Basically the best local linear approximation to the change of the ( linear) function is given by the change in the linear function.

 

[jbriggs444]

I don't think you're getting an exact

If $f(x) = x^2$ then the formula for the slope of a secant line: $\frac{f(x+\Delta x) - f(x)}{\Delta x}$ is exact.

It is exactly $\frac{(x + \Delta x)^2 - x^2}{\Delta x} = \frac{x^2 + 2x\Delta x + \Delta x^2 - x^2}{\Delta x} = \frac{2x\Delta x^2 + \Delta x^2}{\Delta x} = 2x + \Delta x$

That formula is exact and is valid for $\Delta x \ne 0$.

If we define the slope of $f()$ at $x$ to be the limit of the slope of these secant lines as $\Delta x$ approaches zero then we are asking about $\lim_{\Delta x \to 0}\ 2x + \Delta x$.

That limit is $2x$ exactly. [Note the the limit definition carefully never involves what happens when $\Delta x = 0$]

If we do not define the slope of $f()$ at $x$ then we have nothing to talk about.

There may be an exception , both in the definition, result, when we consider the linear change of a line; linear object ( n-dimensional linear object). Here, the approximation to the change is _ equal_ to the change, as , given say a line$f(x)=ax$ ( ignore the constant; just a translation) the limit quotient becomes $\frac {ax+a(\Delta x)-(ax)}{\Delta x}=\frac{a\Delta x}{\Delta x}$ which equals $a$ as long as $\Delta x \neq 0$. Basically the best local linear approximation to the change of the ( linear) function is given by the change in the linear function.

If $f(x) = ax$ then the same process produces a formula for the slope of the secant lines: $a$.

If we define the slope of this new $f()$ at $x$ to be limit of the slopes of these secant lines as $\Delta x$ approaches zero then we are asking about $lim_{\Delta x \to 0}\ a$.

Unsurprisingly, that limit is $a$ exactly.

 

[WWGD]

If $f(x) = x^2$ then the formula for the slope of a secant line: $\frac{f(x+\Delta x) - f(x)}{\Delta x}$ is exact.

It is exactly $\frac{(x + \Delta x)^2 - x^2}{\Delta x} = \frac{x^2 + 2x\Delta x + \Delta x^2 - x^2}{\Delta x} = \frac{2x\Delta x^2 + \Delta x^2}{\Delta x} = 2x + \Delta x$

That formula is exact and is valid for $\Delta x \ne 0$.

If we define the slope of $f()$ at $x$ to be the limit of the slope of these secant lines as $\Delta x$ approaches zero then we are asking about $\lim_{\Delta x \to 0}\ 2x + \Delta x$.

That limit is $2x$ exactly. [Note the the limit definition carefully never involves what happens when $\Delta x = 0$]

If we do not define the slope of $f()$ at $x$ then we have nothing to talk about.


If $f(x) = ax$ then the same process produces a formula for the slope of the secant lines: $a$.

If we define the slope of this new $f()$ at $x$ to be limit of the slopes of these secant lines as $\Delta x$ approaches zero then we are asking about $lim_{\Delta x \to 0}\ a$.

Unsurprisingly, that limit is $a$ exactly.

But for a line, the change is constant,i.e., the limit is constant, not a function of x nor delta x, rather than a function. Unlike in other cases, the linear change in the change near a point is _ equal_ to the change of the function, and constant. In terms of differentials, for linear functions L(x), dL(x)=L(x)dx.

 

[jbriggs444]

But for a line, the change is constant,i.e., the limit is constant, not a function of x nor delta x, rather than a function.

A constant function is still a function.

 

[WWGD]

A constant function is still a function.

My point is the approximation to the change along a linear object is given _ exactly_ , by the change of the linear object, rather than by an approximation along a linear hyperplane. No limits needed. The change in the value of L(x)=ax from x to xo:=x+Deltax can be _ exactly_ determined as a(xo+Delta x)-axo== a*Delta x. It needs no use of limits, unlike the case for nonlinear functions.

 

[WWGD]

Do tell, @PeroK , your specific objections.

 

[WWGD]

Specifically, for, say, f(x)=2x, the change of f(x)=2x, from x=2, to x=2.01, is _ exactly_, independent of any limiting process, equal to 2(2.01)-2(2)= 0.02. You can't do this for , say sinx, nor x^3, etc. , where we estimate the change, rather than compute it.

 

[PeroK]

Do tell, @PeroK , your specific objections.

I've been scrupulously avoiding reacting to your posts with scepticsm and contenting myself by liking those of the man from Maryland.

 

[jbriggs444]

Specifically, for, say, f(x)=2x, the change of f(x)=2x, from x=2, to x=2.01, is _ exactly_, independent of any limiting process, equal to 2(2.01)-2(2)= 0.02. You can't do this for , say sinx, nor x^3, etc. , where we estimate the change, rather than compute it.

So your objection is that (for $\sin x$, $x^3$, etc) that an approximation to the slope changes depending on where you choose to measure the approximation? Yes. So?

Do you conclude that the limit approached by the set of approximations does not exist? That would be false.

Or that the limit is inexact? That would be false. If the limit exists, it is exact and unique.

Or that the notion of the "slope" of a function at a point is inherently undefined. You might argue for that position. But such an argument would be pointless. Mathematics is not about things that lack definitions. It is about things that have definitions.

 

[WWGD]

So your objection is that (for $\sin x$, $x^3$, etc) that an approximation to the slope changes depending on where you choose to measure the approximation? Yes. So?

Do you conclude that the limit approached by the set of approximations does not exist? That would be false.

Or that the limit is inexact? That would be false. If the limit exists, it is exact and unique.

Or that the notion of the "slope" of a function at a point is inherently undefined. You might argue for that position. But such an argument would be pointless. Mathematics is not about things that lack definitions. It is about things that have definitions.

I never said any such thing. I said for a line , one we can find the _ actual_ change, without the use of limits. That can't be done for nonlinear functions. That's a fact.

 

[WWGD]

Anything, @PeroK ? Want to chime in?

 

[jbriggs444]

I never said any such thing. I said for a line , one we can find the _ actual_ change, without the use of limits. That can't be done for nonlinear functions. That's a fact.

What actual change are you measuring?

Do you claim that you are actually measuring the slope of a straight line at a point? Without having first defined the slope of a function at a point.

Do you claim that our calculations for the slope of various secants are not actual or are not exact? Or are you simply pointing out that you can find secants that fail to match the slope of a function at a point. Yes. You can. We all agree about that. We can even find examples like $f(x) = x^3$ where every secant will have positive slope while (we claim that) the slope of $f(x)$ at $x=0$ is zero. Did you have a point to make beyond that?

I definitely respect the choice by @PeroK not to respond. It is difficult to argue about a viewpoint that is deeply held but which is not being coherently stated.

 

[PeroK]

I'm too busy rustling up a chicken curry.

 

[WWGD]

What actual change are you measuring?

Do you claim that you are actually measuring the slope of a straight line at a point? Without having first defined the slope of a function at a point.

Do you claim that our calculations for the slope of various secants are not actual or are not exact? Or are you simply pointing out that you can find secants that fail to match the slope of a function at a point. Yes. You can. We all agree about that. We can even find examples like $f(x) = x^3$ where every secant will have positive slope while (we claim that) the slope of $f(x)$ at $x=0$ is zero. Did you have a point to make beyond that?

I definitely respect the choice by @PeroK not to respond. It is difficult to argue about a viewpoint that is deeply held but which is not being coherently stated.

It's your misframing my reply that is the problem here. Yes, df is the _ approximate_ change along the tangent line/plane. While $\Delta f$ is the _ actual_ change. That is a fact, and not lacking in coherence. Maybe @fresh_42 can chime in.

 

[jbriggs444]

It's your misframing my reply that is the problem here. Yes, df is the _ approximate_ change along the tangent line/plane. While $\Delta f$ is the _ actual_ change. That is a fact, and not lacking in coherence. Maybe @fresh_42 can chime in.

I agree that the first derivative of a function will [often] give rise to a useful linear approximation to the function value in an interval. That is an immediate consequence of Taylor's theorem.

This does not mean that $df$ (or, more correctly, $\frac{df}{dx}$ or $f'(x)$) is only approximate. It is exact.

Formally, $df$ is a piece of notation, not really a number. I assume that you are using it to refer to the first derivative of the function $f()$ at a particular point.