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.

  • $\begingroup$ 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

Hint: Use L'Hopital's Rule.${}$

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.