Why can't the antiderivative of 1/x be found using the limit of the power rule? Which step does this go wrong?
$$\frac{d}{dx} x^n = nx^{n-1}$$
$$∫ \frac{d}{dx} \left(x^{n} \right) dx = ∫ n x^{n-1} dx$$
$$x^n + c = n ∫ x^{n-1} dx$$
$$\frac{x^n}{n} + c = ∫ x^{n-1} dx, n≠0$$
$$\lim_{n→0} \left[ \frac{x^{n}}{n} + c = ∫ x^{n-1} dx \right]$$
However
$$\lim_{n→0} \left[ \frac{x^{n}}{n}+c \right] DNE$$
Therefore
$$ \lim_{n→0} \left[ ∫ x^{ n-1} dx \right] DNE$$
I understand that $∫ x^{-1} dx = \ln(x)$, but why doesn't the limit work?
 A: The main problem is that you cannot do limits of indefinite integrals
If you consider $0<a<b$ and the identity (for $n\neq0$):
$$\int_a^b x^{n-1}\ \mathrm dx = \frac{b^n-a^n}n$$
then if you do the limit for $n→0$ you get
$$\lim_{n\to0}\int_a^b x^{n-1}\ \mathrm dx = \lim_{n\to0}\frac{b^n-a^n}n$$
The limit in the RHS is $\log(b)-\log(a)$ so you really get that
$$\lim_{n→ 0} \int_a^b x^{n-1}\ \mathrm dx = \int_a^b x^{-1}\ \mathrm dx.$$
Note that, in general, you cannot swap arbitrarily the integral with the limit!
A: I have an idea for this...
The step wrong was between steps $3$ and $4$, where, on the LHS I went from "$x^n+c = RHS$" to "$\frac{x^n}{n}+c=RHS"$
The limit definition of $\ln(x)$ is
$$\lim_{h→0} \left[\frac{x^h-1}{h}\right]$$
If I define $c=-1+n\cdot c_1$, we can rewrite steps 3 and on as
$$x^n-1+n\cdot c_1 = n∫x^{n-1}dx $$
$$\frac{x^n-1}{n} +c_1 = ∫x^{n-1}dx, n≠0 $$
$$\lim_{n→0}\left[\frac{x^n-1}{n} +c_1 = ∫x^{n-1}dx \right]$$
$$\lim_{n→0}\left[\frac{x^n-1}{n} +c_1\right] = \ln(x) + c_1$$
Therefore
$$\lim_{n→0}\left[ ∫ x^{n-1} dx \right] = \ln(x) + c_1$$
So essentially, the problem was I wrongly assumed $\frac{c}{n}$ was also a Real constant, and the value of that constant could have made the limit converge.
Please correct me if I've made a mistake.
