Maeckes logo

<    1    >





We call this infinitely small value Δx, and keep in mind that Δx→0. Thus applies

as Δx is neglectable. However, Δx has a value, so you may use it as a divisor. In the further investigation we use the formula

and calculate, with a blind eye, in a first step


That is however wrong. You cannot do that. We try again, in the proper way, and start with n = 0. For every number a ≠ 0 applies a0 = 1. This time the formula gives

It looks quite clear now. The result is the same in both calculations. We continue with n = 1 and find

That is strange, because that is another result from our calculation with the blind eye. Let's try with n = 2 and we see

Once more, now with n = 3 and get

Apparently is (1 + Δx) smaller than (1 + Δx)2 and that is again smaller than (1 + Δx)3. That could be an explanation. But that cannot be true. We have no doubt that

Moreover we just saw that

and thus every exponent, let us just call it n, must of course give 1 as result, because generally applies

What has gone wrong? Alas, here you will notice that infinitely small is different from zero. The big mistake was made from the very beginning, as it should read

because it is really a limit that we handle here. Special rules apply when processing them. You should never apply a limit to a part of a calculation. So, an error was made, because

and that had fatal consequences, as we have seen.


Deutsch   English   Español   Français   Nederlands