I am trying to find out why $(\log(n))^{99} = o(n^{\frac{1}{99}})$. I tried to find the limit as this fraction goes to zero.
$$ \lim_{n \to \infty} \frac{ (\log(n))^{99} }{n^{\frac{1}{99}}} $$
But I'm not sure how I can reduce this expression.
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$\qquad \begin{align} \lim_{x \to \infty} \frac{ (\log(x))^{99} }{x^{\frac{1}{99}}} &= \lim_{x \to \infty} \frac{ (99^2)(\log(x))^{98} }{x^{\frac{1}{99}}} \\ &= \lim_{x \to \infty} \frac{ (99^3) \times 98(\log(x))^{97} }{x^{\frac{1}{99}}} \\ &\vdots \\ &= \lim_{x \to \infty} \frac{ (99^{99})\times 99! }{x^{\frac{1}{99}}} \\ &= 0 \end{align}$
I used L'Hôpital's rule law in each conversion assuming natural logarithm.
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1$\begingroup$ Important to note: this works because $99$ is finite; I have seen people hide $n$ applications of L'Hôpital in $\dots$. Also, be very wary of using the rule in the context of discrete functions; I have seen people (trying to) apply it to $n!$. $\endgroup$ – Raphael♦ Feb 17 '13 at 8:56
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1$\begingroup$ Please consider not reinforcing behaviour that hurts the site. $\endgroup$ – Raphael♦ Feb 17 '13 at 10:18
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$\begingroup$ @Raphael: I think in the case of $n!$, $\Gamma(x)$ or Stirling's approximation could be used to make it suitable for applying L'Hôpital for large $n$. $\endgroup$ – Reza Feb 17 '13 at 23:42
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$\begingroup$ Go ahead and try; not only is the derivation of $\Gamma$ even harder to handle, but it is clear that L'Hôpital can not help for exponential functions: the derivations are (at least) as hard as the original functions. Stirling's formula, on the other hand, is usually a good idea. See also my answer over here. $\endgroup$ – Raphael♦ Feb 18 '13 at 8:46
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Hint: take $99$th root of both sides.
answered Feb 17 '13 at 3:55
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Try a common trick: express both functions as $e^{\dots}$ and compare the exponents; if their ratio tends to $0$ or $\infty$, so does the original quotient (see here for the full rule).
$\frac{(\log n)^{99}}{n^{\frac{1}{99}}} = \frac{e^{99 \cdot \log(\log n)}}{e^{\frac{1}{99} \cdot \log n}}$
