For $a>0$ and $b>0$,

I tried like this

$\lim_{n \to \infty} \frac{ \log^bn }{n^a} = \lim_{n \to \infty}\frac{(\log n)^b }{n^a}$

and then by applying L'Hopital's Rule

$\lim_{n \to \infty} \frac{ b*\log n*1/n }{an^(a-1)}$

Its seems even though I apply L'Hopital's Rule it is giving me indeterminate form.

Any help how above limit becomes 0.

  • $\begingroup$ What are $a,b$? Are they any real number? Because if so, the statement is false. Are they nonnegative numbers? The statement is still false. It may be true if they are both positive real numbers. $\endgroup$ – InterstellarProbe Sep 11 '18 at 21:25
  • 1
    $\begingroup$ It's true when a is positive. b can be any number $\endgroup$ – Jakobian Sep 11 '18 at 21:26
  • $\begingroup$ @Rumpelstiltskin good point. $\endgroup$ – InterstellarProbe Sep 11 '18 at 21:27
  • $\begingroup$ Yes $a>0$ and $b >0$ $\endgroup$ – Joshna Gunturu Sep 11 '18 at 21:27
  • $\begingroup$ The tag refer to limit without l’Hopital. Are interested on that or are you looking for a solution by l’Hopital? $\endgroup$ – gimusi Sep 12 '18 at 7:48

I like to use the integral form of the logarithm: $$\log(x) = \int_1^x \frac 1t \, dt,\ x > 0.$$ It is routine to show that $\displaystyle \lim_{x \to \infty} \frac{\log(x)}{x} = 0$ because for $x > 1$ you have $$0 \le \frac{\log x}{x} = \frac 1x \int_1^{\sqrt x} \frac 1t \, dt + \frac 1x \int_{\sqrt x}^x \frac 1t \, dt \le \frac {(\sqrt x -1)}x + \frac{(x-\sqrt x)}{x \sqrt x} \le \frac{2}{\sqrt x}$$ and can apply the squeeze theorem.

If $a > 0$ then $\displaystyle \lim_{x \to \infty} \frac{\log(x)}{x^a} = 0$ because $x^a \to \infty$ as $x \to \infty$ and $$\frac{\log x}{x^a} = \frac 1a \frac{\log x^a}{x^a} \to 0.$$ If $b > 0$ too then $$\frac{(\log x)^b}{x^a} = \left( \frac{\log x}{x^{a/b}} \right)^b \to 0.$$ If $b \le 0$ then $(\log x)^b \le 1$ as soon as $x \ge e$ and the result is trivial.


Let consider the limit

$$ \lim_{x \to \infty}\frac{ \log^bx }{x^a}$$

with $a,b>0$ and let $x^a=e^y\to \infty$ with $y\to \infty$ therefore

$$ \lim_{x \to \infty}\frac{ \log^bx }{x^a}=\lim_{y \to \infty}\frac{ \log^be^{y/a} }{e^{y}}=\lim_{y \to \infty}\frac1{a^b}\frac{ y^b }{e^{y}}=0$$

which follow by l'Hopital or by the fact that eventually $e^{y}=1+y+\frac12y^2+\ldots \ge y^{2b}$ and then

$$0\le \frac{ y^b }{e^{y}} \le \frac{ y^b }{y^{2b}}=\frac1{ y^b }\to 0$$

Therefore we can conclude that

$$ \lim_{x \to \infty}\frac{ \log^bx }{x^a}=0 \implies \lim_{n \to \infty} \frac{ \log^bn }{n^a} = 0$$

  • $\begingroup$ $\log^bn$ is meaningless, what is it suppose to be ? $\log n^b $ or $(\log n)^b$ $\endgroup$ – Arjang Sep 12 '18 at 8:03
  • $\begingroup$ @Arjang For what I've always seen $\log^bn$ is the standard notation for $(\log n)^b$. $\endgroup$ – gimusi Sep 12 '18 at 8:10
  • $\begingroup$ from what I have seen $\log^bn$ is standard notation for $\log \log \log \cdots \log n $ where $\log$ is applied $b$ number of times. $\endgroup$ – Arjang Sep 12 '18 at 8:13
  • $\begingroup$ @Arjang please refer to math.stackexchange.com/q/150546/505767 $\endgroup$ – gimusi Sep 12 '18 at 8:14

Let's start with the fundamental inequality satisfied by $\log x$ namely $$0\leq\log x\leq x-1,\,x\geq 1\tag{1}$$ Let $c=a/b>0$ and choose $d$ such that $0<d<c$. Replacing $x$ by $x^d$ in $(1)$ we get $$0\leq d\log x\leq x^d-1<x^d,\,x\geq 1$$ which is same as $$0\leq \log x<\frac{x^d} {d}$$ and this implies that $$0\leq \frac{\log x} {x^c} <\frac{1}{dx^{c-d}},\,x\geq 1$$ Noting that $c-d>0$ and taking limits of the above inequality as $x\to \infty $ we get $$\lim_{x\to\infty} \frac{\log x} {x^c} =0$$ via Squeeze Theorem. Raising the above to power $b$ we get $$\lim_{x\to\infty} \frac{(\log x)^{b}} {x^a} =0$$

The fundamental inequality $(1)$ looks very simple, but is one of the defining characteristics of the logarithm function. More formally

Theorem: If $f:\mathbb{R} ^{+} \to \mathbb{R} $ satisfies $f(x) \leq x-1,\,\forall x>0$ and $f(xy) =f(x) +f(y), \, \forall x, y>0$ then $f(x) =\log x$.

  • $\begingroup$ Very nice derivation! How can we justify the foundamental inequality without advanced tools such as derivatives, Taylor? Mayby by $$x=e^y\implies y\le e^y-1$$ and deriving the latter from $e^n\ge n$ which can be shown by induction? $\endgroup$ – gimusi Sep 12 '18 at 8:20
  • $\begingroup$ @gimusi : The inequality is an immediate consequence of any definition of $\log x$ you choose. For the definition based on integral just note that if $1/t\leq 1$ if $t\geq 1$ and integrating this wrt $t$ on $[1,x]$ we get the inequality. For definition $\log x=\lim_{n\to\infty} n(x^{1/n}-1)$ prove the fact that sequence $n(x^{1/n}-1)$ is decreasing. If you define log as inverse of exp then you need to establish $e^x\geq 1+x$ which can be done using any definition of $e^x$. $\endgroup$ – Paramanand Singh Sep 12 '18 at 8:35

Assuming $b>0$:

With L'Hospital's Rule, you get:

$$\dfrac{b\left(\log n\right)^{b-1}\dfrac{1}{n}}{an^{a-1}} = \dfrac{b\log^{b-1} n}{an^a} = \dfrac{b}{a}\cdot \dfrac{\log^{b-1} n}{n^a}$$

If you apply L'Hospital's Rule $\lceil b \rceil$ times, you get:

$$\dfrac{\lceil b \rceil !}{a^{\lceil b \rceil}} \cdot \dfrac{\log^{b-\lceil b \rceil} n}{n^a}$$

This is no longer an indeterminate form.

If $b\le 0$, then it was not an indeterminate form to begin with.


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