Basic d/dx refresher question 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!
 A: 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?
A: 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.)
