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.
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.
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.
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.
$\blacksquare$