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If we have f '(2)=5, does that mean the rate of change 1.999999(infinite number of 9s) to 2 is 5, or to go from 2 to 2.000001 (infinite zeros before 1) is a rate of change of 5?

Thanks!

findingmyway
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  • Side note : $1.99999...$ actually equals $2$. So does $1.000...1$. See this. Moreover, the "rate of change from $x$ to $y$" doesn't mean anything. The derivative of a function at a point is the slope of the curve at that point. – Anthony Aug 27 '20 at 22:15
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    $f'(2) = 5$ just means that the instantaneous rate of change of $f$ at $x = 2$ is $5$. In other words, $\lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} = 5$. – littleO Aug 27 '20 at 22:19
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    I appreciate the response. I remember being taught that the d/dx of a function is what the function's new value would be if x increased by the most marginal amount, which is the same thing as the instantaneous slope you referenced. I just cannot remember if it's the marginal step past the x value or towards the x value. – findingmyway Aug 27 '20 at 22:21
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    Short answer: It's the same value, taking steps "past" $x$ or "towards" $x$. The derivative of a function exists at a point if and only if the left derivative and right derivative both exist and are equal. – Chris Culter Aug 27 '20 at 22:30
  • @ChrisCulter Thanks and understood! This was helpful! :) – findingmyway Aug 27 '20 at 22:30
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    There's no such thing as the "most marginal value". $1.999...$ is equal to $2$. We cannot say that $f'(x) = \frac{f(x + h) - f(x)}{h}$ for some extremely tiny value of $h$. All we can say is that, as $h$ gets closer and closer to $0$, $\frac{f(x + h) - f(x)}{h}$ gets closer and closer to $f'(x)$. It would be worth taking time to understand the concept of a limit more clearly. – littleO Aug 27 '20 at 22:33
  • @AnthonySaint-Criq To be slightly pedantic: $1.000\dots1$ isn't even a number at all – that is not a valid decimal expansion. – diracdeltafunk Aug 27 '20 at 22:43

2 Answers2

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There is no such number as $2.000\ldots 1$, and the number $1.999\ldots$ is exactly equal $2$.

What $f'(2)=5$ means is something else. It means that, by making the interval "around" $2$ (still finite but) small enough, you can make the rate of change of $f$ in that interval arbitrarily close to $5$. In precise mathematical language, however small $\epsilon\gt 0$ you choose, you can find a small $\delta\gt 0$ (which may depend on $\epsilon$) so that, whenever $2-\delta\lt x\lt 2+\delta$, $x\ne2$ you will have $5-\epsilon\lt\frac{f(x)-f(2)}{x-2}\lt 5+\epsilon$.

You may want to be more formal and use symbols such as $\forall$ and $\exists$:

$$(\forall\epsilon\gt 0)(\exists\delta\gt 0)(\forall x)\left(0\lt|x-2|\lt\delta\implies\left|\frac{f(x)-f(2)}{x-2}-5\right|\lt\epsilon\right)$$

That is it, and the whole definition involves only arithmetic with real numbers, no "infinitesimally small" quantities of any sort involved.

Note: I am suspicious whether "infinitesimally small" quantities are even useful for intuition about calculus. They may be truly misleading. There are ways to define them rigorously (look up "non-standard analysis" on the Web), but then you realise dealing with them is harder, rather than easier, than dealing with ordinary real numbers. So - better stay within the realms of the standard real analysis, shall we?

  • I agree that infinitesimals are harder to treat rigorously than the standard real analysis methods. But I think they are extremely useful for intuition in applications like volumes of solids of revolution. Physicists use them correctly all the time. – Ethan Bolker Aug 27 '20 at 23:06
  • @EthanBolker Thanks for the response. As outlined in Jair Taylor's response below, H goes towards zero but doesn't equal zero. This is where my infinitesimally small intuition comes in. The limit as defined is not the slope between x=2 and x=2 but x=2 and x+(an infinitesimal small delta). This is the intuition that guided my original post. – findingmyway Aug 27 '20 at 23:15
  • @EthanBolker That is why I said I was "suspicious" rather than being outright against their use. I am not sure. I know translation of what physicists do to a rigorous mathematical concept is a science in itself. The other day we had a question here from a physicist being baffled by a sleight-of-hand replacement of an ordinary power series with a formal power series when the former diverged. (Apparently physicists call that "renormalisation", and nobody really knows why it works, when it does, and why it doesn't, when it doesn't...) –  Aug 27 '20 at 23:17
  • @findingmyway Indeed, you have a right intuition, however the question sounded like you wanted the actual definition of the derivative (or at least to check how far your intuition was from it) - hope those answers will help. –  Aug 27 '20 at 23:24
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While $1.999\ldots$ is just $2$, and and $2.00\ldots 1$ is not defined as a real number, the concept of limit is similar to this notion in the following way.

By definition, $$f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}.$$

Taking $x = 2$ and $h = .00001$, we have that \begin{align} \frac{f(x+h) - f(x)}{h} = \frac{f(2.00001) - f(2)}{.00001} \tag{1} \end{align} is the rate of change of $f$ between $2$ and $2.00001$. If we take $h = -.00001$, then

$$\frac{f(x+h) - f(x)}{h} = \frac{f(1.99999) - f(2)}{-.00001} = \frac{f(2) - f(1.99999)}{.00001}\tag{2}$$

is the rate of change of $f$ from $1.99999$ to $2$.

Now what the definition of $f'(2) = 5$ says is that as $h\rightarrow 0$, the difference quotient much approach $5$. The limit must be a two-sided limit by definition, so we can take $h \rightarrow 0^-$ or $h \rightarrow 0^+$ and get the same result, $5$. This means that as the number of $0$s in the expression (1) gets large enough, the expression will evaluate to as close to $5$ as you want to go; and similarly, as the number of $9$s in (2) gets large enough, the expression will evaluate to as close to $5$ as you want to go.

(If you want to think in terms of $\epsilon$ and $\delta$: letting $\epsilon > 0$, there is $\delta$ so that $$\big|\frac{f(x+h) - f(x)}{h} - 5\big| < \epsilon$$ for $|h| < \delta$. Then if you choose $n$ so that $10^{-n} < \delta$, you are guaranteed that $$\big|\frac{f(2) - f(1.9999\ldots 9)}{.00\ldots 1} - 5 \big| < \epsilon$$ where $1.9999\ldots 9$ means $1$ followed by $n$ 9s and $.00\ldots 1$ means the decimal is followed by $n-1$ $0$s. And the result is is similar for the right-hand limit.)

Jair Taylor
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