Limits, derivatives, and dividing by zero. [Contradiction in derivative defintions?]

Question

In the limit definition where the denominator is $x - a$, and we take the limit as $x$ approaches $a$, we assume that this denominator is not equal to zero. Where (besides the fact that it is necessary to avoid division by zero) does this assumption come from?

Why is this not in direct contradiction with the definition of continuity, stating that when a function is continuous the limit as $x$ approaches $a$ will equal $f(a)$ so long as the range around $f(a)$ is not infinite.

If $f(x) = x$, and the limit as $x$ approaches $a$ equals $f(a)$, why in the limit definitions are we allowed to make the assumption that the limit of term $x$ as $x$ approaches $a$ will not simply lead to $a - a$, or even more simply the limit as $h$ approaches $0$ will not simply equal $0$? How is computing the limits of the $x$ or $h$ term in the derivative formulas different than computing a limit of a linear function?

Answer 1

The expression we are interested in is:

$\displaystyle \lim_{x\to a} \frac{f(x)-f(a)}{x-a}$

In a comment you talk about "applying the limit laws first". Well, let's try and see how far we get.

Sure, by continuity, and the limit law that "the difference of two limits is the limit of the difference", we have $\displaystyle \lim_{x\to a} (f(x)-f(a)) =0$.

Even easier, we have $\displaystyle \lim_{x\to a} (x-a) =0$.

But now what? It seems as if you want to apply the limit law that "a quotient of limits is the limit of the quotient", or

If $R:=\displaystyle \lim_{x\to a} r(x)$ and $S:=\displaystyle \lim_{x\to a} s(x)$ exist, then the limit $\lim_{x\to a}\dfrac{r(x)}{s(x)}$ also exists and is equal to $R/S$.

Well, if stated like this, this limit law is wrong, and if you look closely in your lecture notes or textbooks, you will see (I hope) that instead, it is only true under a special assumption on $S$. Do you see what is this extra condition?

In other words, thinking you can evaluate $$\displaystyle \lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$ by evaluating $$\displaystyle \frac{\lim_{x\to a} (f(x)-f(a))}{\lim_{x\to a} (x-a)}$$ is exactly the wrong step you are looking for which is "inherently missing some crucial information".

The extra condition on the denominator we would need for that manipulation is exactly not satisfied in your case, and so you cannot apply that "law". The limit of the quotient has to be computed another way, and no contradiction arises.

Answer 2

Continuity of a function involved not just the behavior of a function at a point, but the relationship it has with the point near it so that we know it remains connected there in some meaningful sense. So since we're considering point near but not exactly at that point limits avoid division more as an artifact of this since we aren't actually concerned with the value $f(a)$ unless we want to claim the function is continuous by comparing it's values to that of the function. If it's not defined at that point then sometime we can define it there by the limit, if it exists, and these are called removable discontinuities since we can get rid of them. All notions of continuity will have to deal with the behavior near, but not exactly at the point and the topological definition of continuity reflects this as well. So it's not a cheat so much as a necessary examination of the neighborhood.