Hello. I need to show that $\sqrt n$ grows faster than $(\log n)^{100}$

Question

Is there an easy way to show that $$\lim_{n\to \infty}\frac {(\log n)^{100}}{\sqrt n}=0 $$

Answer 1

Consider the more general limit: $$ \lim_{x\to \infty}\frac{(\log(x))^{100}}{\sqrt{x}} $$ Then use L'Hopital once to get:

$$ \lim_{x\to \infty}\frac{(\log(x))^{100}}{\sqrt{x}}=200\lim_{x\to \infty}\frac{(\log(x))^{99}}{\sqrt{x}} $$ Now repeat sufficient times to get:

$$ \lim_{x\to \infty}\frac{(\log(x))^{100}}{\sqrt{x}}=K\lim_{x\to \infty}\frac{\log(x)}{\sqrt{x}}=0 $$

Hence for any sequence $a_n,\ n=1, ...$ for which $a_n\to \infty$ we have:

$$ \lim_{n\to \infty}\frac{(\log(a_n))^{100}}{\sqrt{a_n}}=0 $$

etc...

Answer 2

Let $m=\log(n)$, then since $e^x\ge1+x\implies e^x\ge\left(1+\frac xk\right)^k$, let $k=101$ $$ \frac{\log(n)^{100}}{\sqrt{n}}=\frac{m^{100}}{e^{m/2}}\le\frac{m^{100}}{\left(1+\frac{m}{202}\right)^{101}}\le\frac{202^{101}}m=\frac{202^{101}}{\log(n)} $$

Answer 3

Hint:

Take $n=e^t$, and you need to show that $e^{t/2}$ grows faster than $t^{100}$.

Or, taking the $100^{th}$ root, $e^{t/200}$ grows faster than $t$.

Or by rescaling, $e^u$ grows faster than $200u$, which is the same as $e^{u}$ growing faster than $u$ (or $u$ faster than $\log(u)$).

Then for all $u>1$

$$\frac{\dfrac{e^{u+1}}{u+1}}{\dfrac{e^{u}}{u}}=e\frac{u}{u+1}>\frac e2$$

and $$\frac{e^u}u>\left(\frac e2\right)^u.$$

Answer 4

More generally, for any $a>0$, $b > 0$, $\lim_{n \to \infty} \frac{(\log(n))^a}{n^b} = 0 $.

There are a number of ways to show this. I like reducing this to a standard form.

First, replacing $n$ by $n^{1/b}$,

$\begin{array}\\ \lim_{n \to \infty} \frac{(\log(n))^a}{n^b} &=\frac{(\log(n^{1/b}))^a}{n}\\ &=\frac{(\log(n)/b)^a}{n}\\ &=\left(\frac{\log(n)}{bn^{1/a}}\right)^a\\ \end{array} $

Then, by replacing $n$ by $n^a$, this becomes

$\begin{array}\\ \lim_{n \to \infty} \frac{(\log(n))^a}{n^b} &=\left(\frac{\log(n)}{bn^{1/a}}\right)^a\\ &=\left(\frac{\log(n^a)}{bn}\right)^a\\ &=\left(\frac{a\log(n)}{bn}\right)^a\\ \end{array} $

So, if we can show that $\lim_{n \to \infty} \frac{\log(n)}{n} =0 $, by working backwards, we can show that $\lim_{n \to \infty} \frac{(\log(n))^a}{n^b} =0 $.

Note that these substitutions can be combined by replacing $n$ by $n^{a/b}$. We get

$\begin{array}\\ \lim_{n \to \infty}\frac{(\log(n))^a}{n^b} &=\lim_{n \to \infty}\frac{(\log(n^{a/b}))^a}{(n^{a/b})^b}\\ &=\lim_{n \to \infty}\frac{((a/b)\log(n))^a}{n^{a}}\\ &=\lim_{n \to \infty}\left(\frac{(a/b)\log(n)}{n}\right)^a\\ \end{array} $

Answer 5

(1).For any $A>0$ we have $\lim_{x\to \infty}(\ln x)/x^A=0.$

Proof: For $x>0$ we have $$\ln x=\int_1^x (1/t)\;dt \quad\text { and }\quad x^A= A(1+ \int_1^x (1/t^{1-A}\;dt\;).$$ Take any $E>0.$ Take $x_0>1$ such that $1/(x_0)^A<E A.$ For $t\geq x_0$ we have $$(1/t)/(1/t^{1-A})= 1/x^A\leq 1/(x_0)^A<E A.$$ $$\text {For } x\geq x_0 \text { we have } \ln x=\ln x_0+\int_{x_0}^t(1/t)\;dt\leq$$ $$\leq \ln x_0+\int_{x_0}^x (E A)(1/t^{1-A})\;dt=$$ $$=E(x^A-(x_0)^A).$$So for $x\geq x_0$ we have $$0<(\ln x)/x^A\leq (-E (x_0)^A+\ln x_0)/x^A+E<(\ln x_0)/x^A+E.$$ From this, since $A>0$ implies $\lim_{x\to \infty}(\ln x_0)/x^A=0,$ we have $$0\leq \lim_{y\to \infty} \sup_{x>y}\;[(\ln x)/x^A]\leq E \quad \text {for every } E>0.$$ $$\text {Therefore }\quad \lim_{x\to \infty}(\ln x)/x^A=0.$$ (2a). For any $B \ne 0$ and $C>0$ let $A=C/|B|.$ We have $A>0$ and we have $$\ln x>1\implies 0\leq |(\ln x)^B/x^C|\leq (\ln x)^{|B|}/x^C=$$ $$((\ln x))/x^A)^{|B|}$$ which tends to zero as $x\to \infty.$

(2b). For $B=0$ and $C>0$ we have,trivally,$x>1\implies (\ln x)^B/x^C=1/x^C$, tending to $0$ as $x\to \infty.$

(3). For $r>0$ we have $\log_r x=(\ln x)/\ln r$ so for $C>0$ we have $|\log_r x)|^B/x^C=|\ln r|^{-B} \cdot |\ln x|^B/x^C$ which tends to $0$ as $x\to \infty.$