Why do we treat $h$ as approaching to $0$ but we also treat it as equal to $0$ in the derivative definition?

Question

If $f(x) = x^2$, use the derivative definition to prove $f'(x) = 2x$. Starting with the difference quotient:

$$\lim_{{h \to 0}} \dfrac{f(x+h)-f(x)}{h}\ = f'(x)$$

Substituting $f(x) = x^2$, we get:

$$\lim_{{h \to 0}} \frac{(x + h)^2 - x^2}{h} = \lim_{{h \to 0}} \frac{2xh + h^2}{h}$$

Factoring out $h$ in the numerator:

$$\lim_{{h \to 0}} \frac{2xh + h^2}{h} = \lim_{{h \to 0}} \frac{h \cdot (2x + h)}{h} = \lim_{{h \to 0}} 2x + h$$

At this point, we want to find the derivative $f'(x)$ by taking the limit as $h$ approaches $0$:

$$\lim_{{h \to 0}} (2x + h) = 2x$$

This is the step that is confusing to me. In my calculus $1$ classes we would usually plug in $h=0$ to get $f'(x)=2x+0=2x$. However, in the earlier step we factored out $h$, and we treated $h$ as approaching $0$ but not exactly equal to $0$ to be able to divide $h/h$. So, if we treat $h$ as equal to $0$ in the last step then wouldn't this mean that we just divided $0/0$ in the step where we had factored?

My only other guess was that we are allowed to divide $h/h$ because we are treating the $h$ in $2x+h$ as a value that is so small and so close to $0$ that we basically just ignore it, and, since it is not actually equal to $0$ this means that we can still do $h/h$. I was not sure if this is what we are doing and maybe in calculus 1 classes we just plug in $h=0$ in the final step to make it easy?

Answer 1

You've proven that some function i.e. $k(h)=\frac{2xh+h^2}{h}$, equals some other function wherever they are defined, i.e. when $h\neq 0$. This "other" function is $g(h)=2x+h$ which, unlike $k$, is also defined at $0$. And continuous. Therefore $$\lim_{h\to 0}k(h)=\lim_{h\to 0}g(h)=g(0)$$ and everything is fine. The last equality follows from continuity of $g$, this is what allows us to "plug in" $0$. In particular you need to be careful about it, plugging in things won't always work, even when the function has value there. Continuity is crucial.

So we've extended domain to prove something about the original domain, that was a very useful trick. It is not important that the original function is not defined at $0$, because we are interested in limit, not in value.

Of course "continuity" is just a term hiding the fact that $$\lim_{h\to 0}(2x+h)=2x$$ which needs to be shown in some other way anyway.

Answer 2

I think you're exactly right about the fishiness of the situation... :)

In particular, yes, if we take the delta-epsilon characterization/definition of "limit", then we'd have to do a bit more song-and-dance to confirm (by that standard) that cancelling $h$ and then later setting it to $0$ does what it appears to do.

And, yes, as you see, it is far easier to just do that cancellation and then set $h=0$ than to justify it, by any standard. Historically, some people did object to the fishiness of this: famous Bishop Berkeley's essay "The Analyst" in which he referred to "ghosts of departed quantities". :)

For calculus just involving polynomials and rational functions, that process of manipulating $h$ as a non-zero thing, and then seemingly have it be $0$ at the end, can be construed as declaring $h$ an "infinitesimal", and then "taking the standard part" (throwing away infinitesimals) at the end. This was implicit, in my opinion, in Leibniz' writing, for example. A. Robinson (and others) made this completely legit in the mid 1960's. J. Keisler did write both a calculus book and instructors' manual taking this viewpoint, not-so-many years ago. Part of his didactic point was to not require the students to have to cope with the necessary model theory. After all, we don't really demand that calculus students master Cauchy's or Dedekind's construction of the real numbers before taking the derivatives of polynomials. :)

(My feeling has always been that calculus works amazingly well, even if we don't know why, and maybe can't explain or justify it.)

Answer 3

For fixed $~x \in \Bbb{R}, ~\displaystyle \lim_{h\to 0} 2x+h = 2x.~$

This assertion, while true, is not proven by assuming that $~h = 0.~$

Instead, the assertion is proven by proving that $$\forall \epsilon > 0, ~\exists \delta > 0 ~\text{such that}~ 0 < |h - 0| < \delta \implies |(2x + h) - 2x| < \epsilon.$$