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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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  • $\begingroup$ HINT L'hospital! $\endgroup$ – andybenji Dec 9 '12 at 9:59
  • $\begingroup$ @andybenji: I can't use it yet! $\endgroup$ – Chan Dec 9 '12 at 10:00
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Your bound $\log x<x-1$ for $\log x$ can be improved: $\int_1^x\frac1tdt<\int_1^x\frac{1}{\sqrt{t}}dt$.

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  • $\begingroup$ When $t>1$, we have $1/t<1/\sqrt{t}$. Integrate both sides from $1$ to $x$, we get the above inequality. $\endgroup$ – user1551 Dec 9 '12 at 10:19
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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 $\lim_{x\to +\infty}\frac{\log^n(x)}{x}$

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