'Proof ' that $\ln(x)$ converges

Question

Where is the flaw in the following 'proof '?

$$\lim_{x \to \infty}\left[\frac{\mathrm{d}}{\mathrm{d}x}\left\{\ln(x)\right\}\right]=\lim_{x \to \infty}\left[\frac{1}{x}\right]=0 \implies\lim_{x \to \infty}[\ln(x)]=\text{constant} \in \mathbb{R},$$ therefore $\ln(x)$ converges to some real number. $\square$

Answer 1

It's not even clear what it means to say that $\lim_{x\to\infty} f(x)$ "is a constant," since it isn't a function of any variable, it's just a limit, if the limit exists.

But consider $f(x)=\sqrt{x}$. We know that $\lim_{x\to\infty} \frac{d}{dx}f(x) = 0$, but is $\lim_{x\to\infty} \sqrt{x}$ a constant?

One thing we can say, if $\lim_{x\to\infty} \frac{d}{dx}f(x) = 0$, is that if $D$ is any constant, then $$\lim_{x\to\infty} \left(f(x+D)-f(x)\right)=0$$

This claim is just nonsense:

If $$\lim_{x \to \infty}\left[\frac{\mathrm{d}}{\mathrm{d}x}\left\{\ln(x)\right\}\right]=0$$ then $$\lim_{x \to \infty}\ln(x)=\lim_{x \to \infty}\left[\int 0\mathrm{d}x\right]=\text{constant}$$

The fundamental theorem of calculus does not even remotely support that. We have:

$$\ln(x)=\int_{1}^x \frac{1}{t}dt$$

Note, there is no $x$ in that expression except on the limits. So:

$$\lim_{x\to\infty} \ln(x)=\int_{1}^\infty \frac{1}{t}dt$$

The indefinite integral:

This expression:

$$\lim_{x\to\infty}\int \frac{1}{x}dx$$

Doesn't actually make sense, because $\int\frac{1}{x}dx$ is not well-defined, and, while it is common to refer it as a function of $x$, $x$ is not a free variable.

A simple example, with no infinities: $$\lim_{x\to 1}\int x\,dx = \lim_{x\to 1} \frac{x^2}{2} = \frac{1}{2}$$ But:

$$\int \left(\lim_{x\to 1}x\right)\,dx=\int dx=x$$

You simply can't swap integrals and limits that way.

Answer 2

Consider

$$\int\lim_{x \to a} \frac{1}{x}\mathrm dx = \frac xa + C \neq \lim_{x \to a}\int \frac{1}{x}\,\mathrm dx = \log(a) + C$$

It should be clear that in general this procedure produces functions of $x$ on the left and constants on the right.

Answer 3

Notice though the derivative of ln(x) converges to $0$ that for any interval $[a,b]$ (for $a,b\in\mathbb{R})$ that the derivate is positive every where, thus though yes the derivative does go to $0$ as you approach infinity. At no point is it actually $0$ any interval of $[0,\infty)$ thus you can't employ the property that if if a function has derivative $0$ everywhere within some interval then it is constant on that interval.

Note also if what you said is true that would imply there would be for some a such $[a,\infty)$ that $e^{x}$ wouldn't be a valid function (since it would just a vertical line at that point)

Answer 4

It appears you are saying that if $$ \lim_{x\to\infty}f(x)=0\implies\lim_{x\to\infty}\int_0^xf(x)\,\mathrm{d}x\text{ converges} $$ in your instance, $f(x)=\frac1x$.

Although this would make analysis of integrals much simpler, it is not true. There are many functions that vanish at infinity whose integral diverges ($\frac1x$ being a good example).

Answer 5

I don't know specifically what about your proof is wrong aside from the non sequitur.

if I wanted to prove the opposite I might look at a higher base that also happens to be an integer, which will be easier to think about... so if

There exists no limit to log10(x) as x approaches infinity, then no limit exists for ln(x) as x approaches infinity.

This is easier for me to mentally prove for a very large x there exists a number n that is the number I need to satisfy 10^n = x