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I saw a video on logarithms saying if there is a limit where $x$ approaches $\pm\infty$ of some fraction, then we can solve by using these rules:

  • If the largest power on the top and bottom are the same, then the limit is the division of the leading coefficients;

  • If the largest power is on the bottom, then the limit is $0$;

  • if the largest power is on the top then there is no horizontal asymptote.

For example, $\lim_{x\to\infty}\frac{4x+2}{2x+8}$ is $\frac{4}{2}$ or $2$ because the powers of $x$ are the same and we divide the leading coefficients.

Another one: $\lim_{x\to-\infty}\frac{x+11}{x^3+16}$ is $0$ because the largest power is on the bottom.

My question: How do I compute the limit $\lim_{x\to\infty}\frac{\log x}{x^{1/k}}$? This is particularly difficult for me because of the arbitrary variable $k$, and the fact that the largest power on top ($1$) is only greater than $\frac{1}{k}$ for all $k > 1$.

So there's a case where they can both be the same (where $k=1$ in which case we divide the leading coefficients which is 1), or the case where the largest power is on top ($k>1$), in which case there is no horizontal asymptote.

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  • $\begingroup$ In general, $\log x$ is much smaller than $x^a$ for any $a>0$, so that limit is zero. The answer below suggests how to prove that. $\endgroup$ – Thomas Andrews Nov 5 '14 at 1:13
  • $\begingroup$ @ThomasAndrews But what if I choose to make $a$ small enough that $x^a$ is less than $\log x$? $\endgroup$ – user6607 Nov 5 '14 at 2:22
  • $\begingroup$ It won't be smaller then $\log x$ for $x$ large enough. It will only be smaller for some $x$. $x^2+2$ is sometimes smaller than $x+10000000$, but $\lim_{x\to\infty} \frac{x+100000000}{x^2+2} = 0$. $\endgroup$ – Thomas Andrews Nov 5 '14 at 2:23
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Hint: Use L'Hopital's Rule.${}$

I.e. $$\lim_{x \to \infty}\frac{\log x}{x^{1/k}}=\lim_{x \to\infty}\frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(x^{1/k})}.$$

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  • $\begingroup$ Thanks. I'm going to learn L'Hoptial's Rule now. :) $\endgroup$ – user6607 Nov 5 '14 at 1:15
  • $\begingroup$ Great! Be very careful with it. For instance, you can't apply that rule to $$\lim_{x\to\infty}\dfrac{1}{\log x}.$$ $\endgroup$ – Vladimir Vargas Nov 5 '14 at 1:21
  • $\begingroup$ Please read here for a very nice proof of that rule. $\endgroup$ – Vladimir Vargas Nov 5 '14 at 1:24
  • $\begingroup$ I have a question about the baby rule: He says that rule isn't good enough to compute $\lim_{x\to0}\frac{1-\cos(2x)}{x^2}$ because $g'(x)=0$. When he says $g'(x)=0$ I think he's referring to the derivative of the denominator. But isn't $2x$ the derivative of $x^2$, not $0$? $\endgroup$ – user6607 Nov 5 '14 at 2:34
  • $\begingroup$ Notice that it says $g'(0)=0$. Indeed $g'(x)=2x$. $\endgroup$ – Vladimir Vargas Nov 5 '14 at 2:38

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