Results 1 to 9 of 9
Like Tree3Thanks
  • 1 Post By Plato
  • 1 Post By Plato
  • 1 Post By Plato

Thread: Using arithmetic of limits

  1. #1
    Junior Member
    Joined
    Nov 2018
    From
    France
    Posts
    40

    Using arithmetic of limits

    Hey.

    I have some exercise in CALCULUS, I haven't succeeded to solve:

    For all x>0 and Natural n we get: $(1+x)^n>\frac{n(n-1)}{2}x^2$

    Prove that if $1<q\in \mathbb{R}$ is some constant value then: $\displaystyle{\lim_{n \to \infty}\frac{n}{q^n}}=0$

    I actually tried to start by the inequality as I know the left side is from the identity of Bernoullis inequality but it led me nowhere...

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,155
    Thanks
    3048
    Awards
    1

    Re: Using arithmetic of limits

    Quote Originally Posted by CStudent View Post
    Prove that if $1<q\in \mathbb{R}$ is some constant value then: $\large\color{blue}{\displaystyle{\lim_{n \to \infty}\frac{n}{q^n}}=0}$
    I actually tried to start by the inequality as I know the left side is from the identity of Bernoullis inequality but it led me nowhere...
    I an confused is that what you want to prove?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2018
    From
    France
    Posts
    40

    Re: Using arithmetic of limits

    Quote Originally Posted by Plato View Post
    I an confused is that what you want to prove?
    Yes
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,155
    Thanks
    3048
    Awards
    1

    Re: Using arithmetic of limits

    Quote Originally Posted by CStudent View Post
    Yes
    So we are put at a disadvantage. We have no idea what theorems and/or axioms/definitions you have to work with.
    This is a fairly standard result but it does come in the sequence of well developed set of theorems.
    Here is an example.
    If $q>1$ then $q=(1+x)$ so
    $\displaystyle \begin{align*}q^n=(1+x)^n&\ge\dfrac{n(n-1)x^2}{2} \\\dfrac{1}{q^n}&\le\dfrac{2}{n(n-1)x^2}\\\dfrac{n}{q^n}&\le\dfrac{2}{(n-1)x^2} \end{align*}$
    Thanks from CStudent
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2018
    From
    France
    Posts
    40

    Re: Using arithmetic of limits

    Quote Originally Posted by Plato View Post
    So we are put at a disadvantage. We have no idea what theorems and/or axioms/definitions you have to work with.
    This is a fairly standard result but it does come in the sequence of well developed set of theorems.
    Here is an example.
    If $q>1$ then $q=(1+x)$ so
    $\displaystyle \begin{align*}q^n=(1+x)^n&\ge\dfrac{n(n-1)x^2}{2} \\\dfrac{1}{q^n}&\le\dfrac{2}{n(n-1)x^2}\\\dfrac{n}{q^n}&\le\dfrac{2}{(n-1)x^2} \end{align*}$
    Hey thank you.
    We have reached to the Cantor's lemma.

    A few questions on your solution:
    How are we sure that n is not equal 1?

    How do we know from this inequality now that the limit of the sequence n/q^n is 0? According to what?
    We know that 2 is a constant sequence so its limit is 2, we know that x^2 is also constant so its limit is x^2, and (n-1) tends to infinity (Do I have to prove it somehow?) - So we have 2 at the nominator and infinity at the denominator - which theorem tells me that yields 0?

    And we need to pose another 0 left to n/q^n to use the sandwich, but how do we do that?

    Thank you!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,155
    Thanks
    3048
    Awards
    1

    Re: Using arithmetic of limits

    Quote Originally Posted by CStudent View Post
    Hey thank you.
    We have reached to the Cantor's lemma.
    A few questions on your solution:
    How are we sure that n is not equal 1?
    How do we know from this inequality now that the limit of the sequence n/q^n is 0? According to what?
    We know that 2 is a constant sequence so its limit is 2, we know that x^2 is also constant so its limit is x^2, and (n-1) tends to infinity (Do I have to prove it somehow?) - So we have 2 at the nominator and infinity at the denominator - which theorem tells me that yields 0?

    And we need to pose another 0 left to n/q^n to use the sandwich, but how do we do that?
    How are we sure that n is not equal 1? $n\to \infty$ so what does it matter?
    It seem that you have answered your own questions.
    $0< \dfrac{n}{q^n}\le\dfrac{2}{(n-1)x^2}\to 0$
    Thanks from CStudent
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2018
    From
    France
    Posts
    40

    Re: Using arithmetic of limits

    Quote Originally Posted by Plato View Post
    How are we sure that n is not equal 1? $n\to \infty$ so what does it matter?
    It seem that you have answered your own questions.
    $0< \dfrac{n}{q^n}\le\dfrac{2}{(n-1)x^2}\to 0$
    Why we can pose another 0 left to n/q^n? What enables this step?
    And how do I explain that the very right fraction is also tending to 0?

    Thank you!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,155
    Thanks
    3048
    Awards
    1

    Re: Using arithmetic of limits

    Quote Originally Posted by CStudent View Post
    Why we can pose another 0 left to n/q^n? What enables this step?
    If you do no understand that $q>0~\&~n\to\infty$ implies that $\dfrac{n}{p^n}>0$ you are wasting time with this.

    Quote Originally Posted by CStudent View Post
    And how do I explain that the very right fraction is also tending to 0?
    This in most elementary of limits: if $A~\&~B$ are non-zero constants then $\displaystyle \mathop {\lim }\limits_{n \to \infty } \frac{A}{{B \cdot n}} = 0$.
    Again if that is not clear to you the don't waste your time.
    Last edited by Plato; Nov 26th 2018 at 07:48 AM.
    Thanks from CStudent
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Nov 2018
    From
    France
    Posts
    40

    Re: Using arithmetic of limits

    Quote Originally Posted by Plato View Post
    If you do no understand that $q>0~\&~n\to\infty$ implies that $\dfrac{n}{p^n}> you are wasting time with this.


    This in most elementary of limits: if $A~\&~B$ are non-zero constants then $\displaystyle \mathop {\lim }\limits_{n \to \infty } \frac{A}{{B \cdot n}} = 0$.
    Again if that is not clear to you the don't waste your time.
    Implies that what?
    Okay thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Arithmetic Progression or Arithmetic Series Problem
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Oct 8th 2009, 01:36 AM
  2. Using limits to find other limits
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Sep 18th 2009, 06:34 PM
  3. Function limits and sequence limits
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Apr 26th 2009, 02:45 PM
  4. arithmetic
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Apr 26th 2009, 11:24 AM
  5. [SOLVED] [SOLVED] Limits. LIMITS!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 25th 2008, 11:41 PM

/mathhelpforum @mathhelpforum