Proof of a generic limit with log and exponential

Question

How do I prove that

$\lim_{n \to \infty} \frac{\log_a n}{n^b}=0$

Answer 1

Consider for $b>0$ the corresponding limit in $\mathbb{R}$

$$\lim_{x \to +\infty} \frac{\log_a x}{x^b}$$

For $a>1$ set $x=a^y \to +\infty$

$$\lim_{x \to +\infty} \frac{\log_a x}{x^b}=\lim_{y \to +\infty} \frac{\log_a a^y}{a^{by}}=\lim_{y \to +\infty} \frac{y}{a^{by}}=0$$

indeed $\exists y_M$ such that $\forall y\ge y_M$

$$a^{by}\geq y^2$$

and then

$$\frac{y}{a^{by}}\le\frac{y}{y^2}=\frac1y\to0$$

For $0<a<1$ set $c=\frac1a$ and $x=\left(\frac1a\right)^y=c^y \to +\infty$

$$\lim_{x \to +\infty} \frac{\log_a x}{x^b}=\lim_{y \to +\infty} \frac{y\log_a \frac1a}{c^{by}}=\lim_{y \to +\infty} \frac{-y}{c^{by}}=0$$

Thus

$$\lim_{x \to +\infty} \frac{\log_a x}{x^b}=0 \implies \lim_{n \to +\infty} \frac{\log_a n}{n^b}=0$$

Answer 2

Just change basis to basis $\;e\;$ , say (and assume $\;b>0\;$, as commented):

$$\frac{\log_an}{n^b}=\frac{\frac{\log n}{\log b}}{e^{b\log n}}=\frac{\log n}{\log b\cdot\left(e^b\right)^{\log n}}\xrightarrow[n\to\infty]{}0$$

Answer 3

If $a>1$, you could substitute $n=a^t$ and the limit becomes

$$\lim_{t\to\infty} \frac{t}{a^{bt}}.$$

If $0<a<1$, use $n=a^{-t}.$ Now use L'hospital (or series) if that's allowed, or perhaps a concavity argument.

Answer 4

You can make the limit non-generic by setting $n=\sqrt[b]m$ so that

$$\frac{\log_a n}{n^b}=\frac1{b\log a}\frac{\log m}m$$ and the constants can be factored out. This is undefined when $b=0$ or $a\le0$ or $a=1$.

For $b<0$, the limit relates to $m\to0^+$ and does not exist ($\color{red}{\pm\infty}$ depending on the sign of $\log a$ if you prefer).

And for $b>0$, it is $\color{red}0$, because by the ratio test with $m=e^k$,

$$\frac{\dfrac{\log e^{k+1}}{e^{k+1}}}{\dfrac{\log e^k}{e^k}}=\frac{k+1}{ek}<1.$$

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These results extend to the case of

$$\lim_{n\to\infty}\frac{(\log_an)^c}{n^b},$$ as it can be written

$$\lim_{n\to\infty}\left(\frac{\log_an}{n^{b/c}}\right)^c$$ and is the $c^{th}$ power of the previous limit with $b\to b/c$.

Answer 5

First note that $\log_{a} n=\log n/\log a$ so it is sufficient to prove that $n^{-b} \log n\to 0$ as $n\to\infty$. Next note that $b>0$ and hence $n^b\to\infty$. And $n^{-b} \log n=(1/b)n^{-b} \log n^{b} $ therefore it is sufficient to prove that $(\log x) /x\to 0$ as $x\to\infty$ where $x$ is a real variable.

Next we can use the defining inequality satisfied by $\log x$ given below $$\log x\leq x-1<x$$ and replacing $x$ by $\sqrt{x} $ we get $$\log x<2\sqrt{x}$$ Thus if $x>1$ then $$0<\log x<2\sqrt{x}$$ or $$0<\frac {\log x} {x} <\frac{2}{\sqrt{x}}$$ and applying Squeeze Theorem we get the desired result $$\lim_{x\to \infty} \frac{\log x} {x} =0$$

If one carefully notes the above proof, it will be clear that only two properties of the logarithm function have been used here:

• $\log(xy) =\log x+\log y,\, x,y>0$
• $\log x\leq x-1,\, x>0$

These two properties characterize the logarithm function uniquely and all the other properties of logarithm function can be derived using these properties.