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Prove $\lim_{x\to \infty} \dfrac{\log(x)}{x} = 0$ and $\lim_{x\to \infty} \dfrac{\log(x)}{x^n} = 0$

From the definition of $\log(x)$, $\log(x) = \int_1^x \dfrac{1}{t} dt$ Since $1$ is the $\sup\{f(t) : m_{i-1} \leq t \leq m_i\}$, it follows that $ \int_1^x \dfrac{1}{t} dt \leq U(f, P) < x - 1$ So $\log(x) < x - 1 < x \Rightarrow \dfrac{\log(x)}{x} < 1$ Up to here, I was stuck. I'm thinking of using Squeeze theorem with $\dfrac{1}{x} \leq \dfrac{\log(x)}{x} \leq g(x)$ for some valid $g(x)$. But I couldn't think of such a $g(x)$.
Could anyone share me some ideas on how to solve this problem?

Note L'Hospital's rule is not allowed.

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    @andybenji: I can't use it yet!2012-12-09

4 Answers 4

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Your bound $\log x for $\log x$ can be improved: $\int_1^x\frac1tdt<\int_1^x\frac{1}{\sqrt{t}}dt$.

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    When t>1, we have 1/t<1/\sqrt{t}. Integrate both sides from $1$ to $x$, we get the above inequality.2012-12-09
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Try $g(x)=\sqrt x$ all you need is to show that $\log(x)<\sqrt x$ for $x$ sufficiently large

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$\lim_{x\to \infty} \dfrac{\log(x)}{x^n}$ is of the form $\frac {\infty}{\infty}$ if $n>0$

So, applying L'Hospital's Rule, $\lim_{x\to \infty} \dfrac{\log(x)}{x^n}=\lim_{x\to \infty}\frac{\frac1x}{nx^{n-1}}=\lim_{x\to \infty}\frac1{nx^n}=0$ if $n>0$

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By the First Fundumental Theorem of Calculus, $\log^{\prime}(x)=\frac{1}{x}$ for $x>0$. Observe $\lim_{x\to +\infty}\log x=+\infty$ and so by De L'Hospital Rule, $\lim_{x\to +\infty}\frac{\log(x)}{x}=\lim_{x\to +\infty}\frac{\log^{\prime}(x)}{x^{\prime}}\lim_{x\to +\infty}\frac{\frac1x}{1}=0$ Use the same method for $\lim_{x\to +\infty}\frac{\log^n(x)}{x}$