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Let $ f(x)=\ln(x)$ then as we take the limit to $\infty$,

$$\lim_{x\to \infty} f(x)\to \frac{d}{dx}= \frac{1}{x} \to 0.$$

So if $\dfrac{d}{dx}$ is always getting smaller how does $\lim_{x\to \infty} f(x) \to \infty$

Since we end up adding only decimals intuitively I expected the limit to approach a real number.

amWhy
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Sean D
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    What about $\sum_{n=1}^\infty\frac1n=\infty$? It's the same situation, since $\lim_{n\to\infty}\frac1n=0$. – José Carlos Santos Jun 17 '22 at 11:01
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    You mean - how can a function grow slower and slower and still reach infinity? Well, sure it can. There are many such functions: $x\to\sqrt{x}$ comes to mind. Maybe you can look at more discrete data: sequences. The sequence $1,2,2,3,3,3,3,\underbrace{4,4,\ldots,4}8,\underbrace{5,5,\ldots,5}{16},\ldots$ obviously goes to infinity and grows as fast as $\log_2 n$... –  Jun 17 '22 at 11:08
  • $\frac{d}{dx}= \frac{1}{x}$? You should write $\frac{df}{dx}= \frac{1}{x}$ – jjagmath Jun 17 '22 at 15:24

1 Answers1

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To make your intuition precise we can write $f(x) = f(1) + \int_1^x f'(x)dx$.

Meaning we start at $1$ and add up all the tiny changes between there and $x$ to get $f(x)$. But in this case

$$f(0) = f(1)+ \int_1^0 f'(x)dx = -\int_0^1 \frac{1}{x}dx = -\infty$$

To see the integral diverges compare it to the series $\sum_{n=1}^\infty \frac{1}{n}$ which diverges by this proof.

$$\sum_{n=1}^\infty \frac{1}{n} \simeq 1 + \sum_{n=2}^4\frac{1}{n} + \sum_{n=4}^8\frac{1}{n}+ \ldots + + \sum_{n=2^k}^{2^{k+1}}\frac{1}{n}+ \ldots$$

Exercise: Of course the above counts $4,8,16,\ldots$ twice each. Fix it.

Exercise: Show each sum on the right is greater than $1/2$.

Daron
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