AnswerBun.com

How to prove that $lim_{xtoinfty}frac{(log_2 x)^3}{x^n}=0$

Mathematics Asked by Shiran Shaharabani on December 19, 2020

I need help with proving the following $$lim_{xtoinfty}frac{left({log}_2xright)^3}{x^n}=0 , quad forall n>0.$$

I used wolframalpha and got it

enter image description here

and I had kind of an intuition because of the limit $frac{ln(x)}x$ , and still I have no idea how to formaly proof it.

Will appreciate any help!

4 Answers

Using the hint of @Kavi Rama Murthy, note that $$f(x)to 0 implies f^{3}(x)to 0$$ In your problem, you can define that $f(x):=frac{log_{2}(x)}{x^{n/3}}$ so, by L'Hospital's Rule you can find that $$frac{log_{2}(x)}{x^{n/3}} to 0 quad text{as} quad x to infty$$ So, by the hint and since that $f^{3}(x)=frac{log_{2}^{3}(x)}{x^{n}}$ so $$frac{log_{2}^{3}(x)}{x^{n}} to 0 quad text{as} quad xto infty$$ this's true for all $n>0$.

Correct answer by Александр Пальма on December 19, 2020

If $K,L$ are positive and $B>1$ then, with $C=1/ln B$, we have (for $x>1$)$$frac {(log_Bx)^K}{x^L}=frac {(Cln x)^K}{x^L}=frac {(Cln x)^K}{(x^{L/K})^K}=$$ $$=frac {(,C(K/L)ln (x^{L/K}),)^K}{(x^{L/K})^K}=$$ $$=C^K(K/L)^Kleft(frac {ln (x^{L/K})}{x^{L/K}}right)^K.$$

Answered by DanielWainfleet on December 19, 2020

Here is a direct way without L'Hospital:

Substituting $x=e^t$ and using $log_2 x= frac{ln x}{ln 2}$ you have

$$lim_{xto infty} frac{(log_2 x)^3}{x^n} =frac 1{ln^3 2}lim_{tto infty}frac{t^3}{e^{nt}}$$

Now, since $e^u = sum_{k=0}^{infty}frac{u^k}{k!}$, you have $e^u > frac{u^4}{4!}$ for any $u>0$. Hence,

$$0leq frac{t^3}{e^{nt}} < frac{t^3}{frac{(nt)^4}{4!}}=frac{4!}{n^4}cdot frac 1t stackrel{ttoinfty}{longrightarrow}0$$

Answered by trancelocation on December 19, 2020

HINT

$$lim_{xrightarrowinfty}frac{left({log}_2{x}right)^3}{x^n}=lim_{xrightarrowinfty}frac{tleft(ln{x}right)^3}{x^n}$$ where $t=ln^32$ $$lim_{xrightarrowinfty}frac{tleft(1-frac1xright)^3}{x^n}lelim_{xrightarrowinfty}frac{tleft(ln xright)^3}{x^n}lelim_{xrightarrowinfty}frac{tleft(x-1right)^3}{x^n}$$

$$to0lelim_{xrightarrowinfty}frac{tleft(ln xright)^3}{x^n}leto0$$

Now the limit is obvious from squeeze theorem.

For the inequality, refer this

Here's a convincing graph: enter image description here

Answered by DatBoi on December 19, 2020

Add your own answers!

Related Questions

How to compute $sum_{n=1}^infty{frac{n}{(2n+1)!}}$?

3  Asked on September 18, 2020 by samuel-a-morales

   

Rational Roots (with Lots of Square Roots!)

1  Asked on September 16, 2020 by fleccerd

   

Is a relation that is purely reflexive also symmetric?

1  Asked on September 13, 2020 by paul-j

 

$3^{123} mod 100$

4  Asked on September 13, 2020 by global05

       

Totally order turing machines by halting

1  Asked on September 10, 2020 by donald-hobson

   

Density of Tensor Products

0  Asked on September 8, 2020 by jacob-denson

   

Ask a Question

Get help from others!

© 2022 AnswerBun.com. All rights reserved. Sites we Love: PCI Database, MenuIva, UKBizDB, Menu Kuliner, Sharing RPP