Question

I have a Calculus question concerning the limit of a expression involving logarithm; details in the reply. I have a Calculus question concerning the limit of a expression involving logarithm; details in the reply. @Mathematics

Answer 1

I must show that
\[\lim_{x\to +\infty} \frac{\ln x}{x} = 0 \]
by knowing that
\[ \int_{1}^{x} \left( \frac{1}{\sqrt{t}} dt \right) \geq \int_{1}^{x} \left( \frac{1}{t} dt \right) \]
Evaluating the integrals above I obtain:
\[ 2\sqrt{x} - 2 \geq \ln x \]
But I don't know how to continue from here.

Answer 2

use L hopitals rule since you obtain infinity over infinity from step 1

Answer 3

what you did just shows a true falsse statement

Answer 4

I should do it without using L'Hôpital's rule.

Answer 5

\[0\le\lim_{x\to +\infty} \frac{\ln x}{x} \le \lim_{x\to +\infty} \frac{2\sqrt{x}-2}{x}=\cdots \]

Answer 6

0

Answer 7

is it true?

Answer 8

I think I get it now. By Squeeze theorem, right?
\[\lim_{x\to +\infty}\frac{2\sqrt{x} - 2}{x}=0\]
Then, by Squeeze theorem, \[\lim_{x\to+\infty}\frac{\ln x}{x}=0\]

Answer 9

yes

Answer 10

\[\lim_{x\to+\infty}\frac{2\sqrt{x} - 2}{x} = \lim_{x\to+\infty}\frac{2\frac{1}{\sqrt{x}} - \frac{2}{x}}{1}=0\]