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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?

Farkle Griffen
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    You are dividing by $0$... – alvoi Dec 08 '22 at 16:30
  • @alvoi, yes that's what makes $x^n/n$ divergent, but if $x^n/n = ∫x^{n-1}dx$, their limits should be the same. Or I suppose you can do it without the division, and you get $\lim_{n→0} x^n = \lim_{n→0} n ∫ x^{n-1}dx ⇒ 1=0$ – Farkle Griffen Dec 08 '22 at 16:39
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    In a way, you can since $\ln(x)=\lim_{n\to0}\frac{x^n-1}{n}$ – Тyma Gaidash Dec 08 '22 at 17:05
  • I think the real issue here is that... whenever you take the indefinite integral of some function (i.e. the integrand), the answer represents a family or set of functions whose derivative is the integrand. However, this does not imply that every function in that family has the same limit, as the OP assumes by taking the limit of both sides of the equation. We are only assured that every function in the family has the same derivative. – RyRy the Fly Guy Dec 08 '22 at 18:40

2 Answers2

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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!

alvoi
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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.

Farkle Griffen
  • 111
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    I think the real issue here is that... whenever you take the indefinite integral of some function (i.e. the integrand), the answer represents a family or set of functions whose derivative is the integrand. However, this does not imply that every function in that family has the same limit, as the OP assumes by taking the limit of both sides of the equation. We are only assured that every function in the family has the same derivative. – RyRy the Fly Guy Dec 08 '22 at 18:41