# Can these rules be used to solve this logarithm?

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.

• 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. – Thomas Andrews Nov 5 '14 at 1:13
• @ThomasAndrews But what if I choose to make $a$ small enough that $x^a$ is less than $\log x$? – user6607 Nov 5 '14 at 2:22
• 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 $\lim_{x\to\infty} \frac{x+100000000}{x^2+2} = 0$. – Thomas Andrews Nov 5 '14 at 2:23

## 1 Answer

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

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

• Thanks. I'm going to learn L'Hoptial's Rule now. :) – user6607 Nov 5 '14 at 1:15
• Great! Be very careful with it. For instance, you can't apply that rule to $$\lim_{x\to\infty}\dfrac{1}{\log x}.$$ – Vladimir Vargas Nov 5 '14 at 1:21
• Please read here for a very nice proof of that rule. – Vladimir Vargas Nov 5 '14 at 1:24
• 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 $0$? – user6607 Nov 5 '14 at 2:34
• Notice that it says $g'(0)=0$. Indeed $g'(x)=2x$. – Vladimir Vargas Nov 5 '14 at 2:38