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What would be the value of $\displaystyle\lim_{x\to\infty}\dfrac{x}{\log x}$? Is it infinity or some constant value?

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  • $\begingroup$ It is infinity. $\endgroup$ – Jika May 21 '14 at 18:33
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    $\begingroup$ Infinity, but why? You can use L'Hôpital's rule. $\endgroup$ – Peter Woolfitt May 21 '14 at 18:33
  • $\begingroup$ Because $x\gg\log x$ as $x$ approaches infinity. $\endgroup$ – Jika May 21 '14 at 18:34
  • $\begingroup$ @GammaFunction Thanks. I meant that $x\in \Omega(\log x)$ $\endgroup$ – Jika May 21 '14 at 18:38
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L'Hopital's rule gives answers but not insight.

Look at this: $$ \begin{array}{c|c} x & \log_2 x \\ \hline 1 & 0 \\ 2 & 1 \\ 4 & 2 \\ 8 & 3 \\ 16 & 4 \\ 32 & 5 \\ 64 & 6 \\ 128 & 7 \\ 256 & 8 \\ 512 & 9 \end{array} $$ As $x\to\infty$, what happens to the ratio of the number in the left column to the number in the right column? That's the answer to the question you posed.

(A moment's thought should tell you why the answer will be the same if the base of the logarithms is any number besides $2$, as long as it's $>1$.)

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By L'Hospital's Rule, we have that $\lim\limits_{x\to \infty}$ $x \over \log(x)$ =$\lim\limits_{x\to \infty}$ $1 \over {1/x}$ = $\lim\limits_{x\to \infty} x$ = $\infty$

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    $\begingroup$ L'Hopital's rule gives answers but not insight. See my posted answer. $\endgroup$ – Michael Hardy May 21 '14 at 18:54
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Consider the change of variables $x=e^y$, here we can consider:

$$\lim_{x\to\infty}\frac{x}{\ln x}=\lim_{y\to\infty}\frac{e^y}{\ln e^y}=\lim_{y\to\infty}\frac{e^y}{y}=\infty$$

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  • $\begingroup$ And how do you justify the last limit? $\endgroup$ – lhf May 21 '14 at 19:29
  • $\begingroup$ @lhf by the fact that exponential is much larger than polynmial, but to put forth more rigour: $\large\lim_{y\to\infty}\frac{e^y}{y}=\lim_{y\to\infty}\frac{1+y+y^2/2!+y^3/3!+...}{y}$ $=\lim_{y\to\infty}\frac{1}{y}+1+y/2!+y^2/3!+...=\infty$ $\endgroup$ – Ellya May 21 '14 at 19:38
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    $\begingroup$ In order to avoid complications with infinite sums, you could use the weaker result $e^{y} > 1+y+y^{2}/2$. $\endgroup$ – madprob May 21 '14 at 20:07
  • $\begingroup$ @madprob: It is useful! $\endgroup$ – Paul May 22 '14 at 12:14

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