How limit $\lim_{n \to \infty} \frac{ \log^bn }{n^a} = 0$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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

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

Answer 4

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 $$\lim_{x\to\infty} \frac{(\log x)^{b}} {x^a} =0$$

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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 $f(x) =\log x$.