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I'm not sure how to prove the following formula:

$$\lim_{x \to \infty}\frac{\log x}{\sqrt{x}} = 0$$

I have understood it graphically, and I want to learn the mathematical prove of this. I have come up with using Taylor expansion of log, but are there any other methods to prove this?

Thank you.

Notation1 I am also wondering whether the methods used in order to prove $$\lim_{x \to \infty}\frac{\log x}{x} = 0$$ can be applied here.

Notations2 Although there have been already a similar question (the proof of more generalized case), I want to start from learn the proof of the easier example, that is, this case.

XYJ
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    Paging Dr. l'Hopital... – Sean Roberson Dec 28 '22 at 02:27
  • Thank you for your answer. L'Hospital's Rule is surely a very strong method in calculation of limits in the form of fraction. – XYJ Dec 28 '22 at 02:37
  • @Ѕᴀᴀᴅ Thank you for your comment. The post is surely related to my question, and it is generalization of my question. However, I feel it is also important to start from understanding the easier case and after that move onto proofs of more generalized claims, especially for beginners. – XYJ Dec 28 '22 at 03:50
  • @Ѕᴀᴀᴅ Actually, I have searched regarding my question before I posted this question. However, I could not find the site you mentioned probably because I only search for ** log x / sqrt **. Could you please give me some advice how to search related questions effectively. (I think this is difficult especially the question is related to math or expressed by some formula.) Thank you. – XYJ Dec 28 '22 at 04:52

2 Answers2

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Let $u= \sqrt x$. Then we want the limit

$$\lim_{u \to \infty} \frac{\log (u^2)}{u}=2\lim_{u \to \infty} \frac{\log u}{u}$$

Can you go on from here ?

leonbloy
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    Thank you for your answer. I didn't come up with using variable conversion, but it's very nice to return my question to $\frac{\log x}{x}$. – XYJ Dec 28 '22 at 02:39
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Use L'Hopital's rule: $$\lim_{k\rightarrow\infty}\frac{\log x}{\sqrt{x}}=2\lim_{k\rightarrow\infty}\frac{1}{\sqrt{x}}$$Can you take it on from here?

Kamal Saleh
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    Thank you for your answer. I realized that L'Hospital's rule is really useful. I want to consider whether L'Hospital's rule can be applied when calculating limits. – XYJ Dec 28 '22 at 02:42
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    By the way, is it the following formula? $$\lim_{k\rightarrow\infty}\frac{\log x}{\sqrt{x}}=2\lim_{k\rightarrow\infty}\frac{1}{\sqrt{x}}$$ (no minus in RHS) – XYJ Dec 28 '22 at 02:43
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    @XYJ No the coefficient is negative, but it actually doesn't really matter if you know the answer. – Kamal Saleh Dec 28 '22 at 03:07
  • Thank you for your comment. My calculation is following $$\lim_{k \to \infty}\frac{\log{x}}{\sqrt{x}} = \lim_{k \to \infty}\frac{\frac{d}{dx}\log{x}}{\frac{d}{dx}\sqrt{x}} = \lim_{k \to \infty}\frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}}$$ Could you please tell me where I mistake? Thank you. – XYJ Dec 28 '22 at 03:32
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    @XYJ Oh wait I'm wrong, sorry about that. – Kamal Saleh Dec 28 '22 at 03:34
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    Thank you for your reply. It's no problem. It is really hard to completely prevent miscalculation. – XYJ Dec 28 '22 at 03:39