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Given: $a\in\mathbb{R}$ constant how do I show that $\mathop {\lim }\limits_{n\to \infty}\frac{cos(n)+a}{\sqrt n}=0 $

Thanks.

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2 Answers 2

up vote 4 down vote accepted

Note:

$\ \ 1$) The numerator is bounded in absolute value by some number $M$,

and,

$\ \ $2) the denominator tends to infinity as $n$ tends to infinity.

So, from 1) the inequality $$-{M\over \sqrt n}\le {\cos(n)+a\over\sqrt n}\le {M\over \sqrt n},$$ holds; and then you can find the limit using the Squeeze Theorem.

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@Anonymous: David wrote both sequences. Above. In his answer. –  martini Mar 28 '12 at 13:07
    
@Anonymous Prove that $\sqrt n$ increases to $\infty$ as $n$ tends to infinity (that is, show that 2) holds). Do you need more detail here? What results do you have available that you can use? –  David Mitra Mar 28 '12 at 14:47
    
Firstly, I really want to thank you for your great help. secondly I'm writing the proof and as soon as I'm done with it I'll write what I did and hopefully it'll be what you meant. –  Anonymous Mar 28 '12 at 14:51
    
OK, I think I got it. here is what I've done. I defined M=|a|+1 and m=a-1 therfore the inequality you wrote in the answer I canged the -M to m and then proved that the limit of any sequence of the type $c\over \sqrt n$ where c is a constant in R equals to 0. Then finished using the Squeeze Theorem. What do you think about it? –  Anonymous Mar 28 '12 at 15:01
    
@Anonymous Looks perfect :) (You did show that the limit of ${c\over\sqrt n}$ as $n$ tends to infinity is 0, of course? One other thing, you could have taken $M=a+1$; that would make the proof a bit nicer.) –  David Mitra Mar 28 '12 at 15:06

By definition approach.

Recalling definition of limit of sequence,

$$\lim_{n\to\infty} a_n=L \Longleftrightarrow \forall\epsilon>0,\exists n_0\in\mathbb{N}, \forall n\geq n_0:|a_n-L|<\epsilon$$

We need to prove that,

$$\mathop {\lim }\limits_{n\to \infty}\frac{\cos(n)+a}{\sqrt n}=0$$

Proof (With $a$>0)

Let $\epsilon>0$ be given. We need to find $n(\epsilon)\in\mathbb{N}$, such that for all $n\geq n_0$ we'll have $${\huge{|}}\frac{\cos(n)+a}{\sqrt n}-0{\huge{|}}={\huge{|}}\frac{\cos(n)+a}{\sqrt n}{\huge{|}}<\epsilon$$

We know, that $\cos(x)$ is bounded by $1$ from above, therefore we can write the following:

$${\huge{|}}\frac{\cos(n)+a}{\sqrt n}{\huge{|}}<{\huge{|}}\frac{1+a}{\sqrt n}{\huge{|}}$$

Solving the inequality:

$${\huge{|}}\frac{1+a}{\sqrt n}{\huge{|}}<\epsilon$$

We'll get this result:

$$n>\frac{(a+1)^2}{{\epsilon}^2}$$.

Therefor, if we take $n_0>\frac{(a+1)^2}{{\epsilon}^2}$, we'll get, ${\huge{|}}\frac{\cos(n)+a}{\sqrt n}{\huge{|}}<\epsilon$.

Similarly, constructing a proof for the case in which $a<0$, we end.

$\blacksquare$

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2  
You ought to write $1 + |a|$ instead of $|1 + a|$ and get $(1 + |a|)^2$ etc. –  Aryabhata Mar 28 '12 at 17:40

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