Show $\underset{n\to \infty}{\lim} \frac{2+\sin(n)}{3-2\cos^2(n)}$ doesn't exist

Question

How to prove for $n \in \mathbb{N}$ that: $$\underset{n\to \infty}{\lim} \frac{2+\sin(n)}{3-2\cos^2(n)}$$ does not exist? Without using functions.

Answer 1

Let $n=2\pi k$, where $k\rightarrow+\infty$ and $k\in\mathbb Z$. Thus, $f(n)=2$.

Let $n=\frac{\pi}{2}+2\pi k$, where $k\rightarrow+\infty$ and $k\in\mathbb Z$. Thus, $f(n)=3$,

which says that the limit does not exist.

If you mean that $n\in\mathbb N$ then since $$f(n)=\frac{2+\sin{n}}{1+2\sin^2n}$$ and $[-1,1]$ is a set of all limit points of $a_n=\sin{n}$, the limit does not exist.

Answer 2

Consider the following two integer sequences: $$ \alpha_n=\lfloor{2n\pi}\rfloor-1, \quad \beta_n=\lfloor{2n\pi}\rfloor+1, $$ where by $\lfloor{x}\rfloor$ I denote the integer part of real number $x$. We have: $$ \sin\alpha_n=\sin(\lfloor{2n\pi}\rfloor-2n\pi-1) $$ and the quantity in parentheses on the right hand side is comprised between $-2$ and $-1$. It follows that $$ -1<\sin\alpha_n<-\sin1. $$ By an analogous reasoning one also finds $$ 0<\sin\beta_n<\sin1. $$ Your sequence can be written as $\displaystyle{2+\sin n\over3+2\sin^2n}$ and from the above inequalities it follows that $$ {2+\sin\alpha_n\over3+2\sin^2\alpha_n}<{2-\sin1\over3+2\sin^21}\approx0.262 \quad\hbox{and}\quad {2+\sin\beta_n\over3+2\sin^2\beta_n}>{2\over3+2\sin^21}\approx0.643. $$ These two extracted sequences cannot thus converge to the same limit, which proves that $\displaystyle\lim\limits_{n\to\infty}{2+\sin n\over3+2\sin^2n}$ cannot exist.

Answer 3

If you know the result that $(\sin n)_n$ is dense in $[-1,1]$, then you can extract two subsequences from your sequence, one with limit $2$, the other with limit $3$.

If you don't, then it may be a good idea to start with the demonstration of this result.

Here is a demonstration of the first fact : consider $G=\left\{n+m\pi,(n,m)\in{\mathbb Z}^2\right\}$. This is a subgroup of $\mathbb R$, which is dense : for every real number $r$, and every $\epsilon>0$, there is a number $a$ in $G$ such that $|r-a|<\epsilon$.

To prove this, consider $\alpha=G\cap{\mathbb R}^{\ast+}$. If $\alpha>0$, it is simple to prove that $G=a{\mathbb Z}$, which means that $\pi=m\alpha$, $1=n\alpha$, so $n\pi-m=0$, and $\pi$ is rational, which is know to be false (there are some more details to complete to finish the demonstration, you can search Google for "subgroups of $\mathbb R$" for further informations).

Now let $c\in[-1,1]$ be a real, we can prove that for every $\epsilon>0$, there are infinitely many $n\in\mathbb N$ such that $[\sin n-c|<\epsilon$. I let you walk threw the details, which would fill another page :-)

If there infinitely such $n$, then it means you can construct an increasing sequence $(\phi_n)$ of integers such that $\sin(\phi_n)\xrightarrow[n\to\infty]{}c$.

For your sequence, first choose $c=0$, the $c=1$...

Answer 4

Consider the function $f(x)=\frac{2+sin(x)}{3-2\cos^2(x)}$. As both the numerator and denominator of $f$ is continuous and the denominator of $f$ is never zero for any $x\in\mathbb{R}$, we see that the function is continuous. As $f(2k\pi)=2$ and $f\left(\frac{\pi}{2}+2k\pi\right)=3$ for all $k\in\mathbb{Z}$, we can say that for every $\epsilon>0$ we can find $\delta_k,\delta_k'>0$ such that $|f(x)-3|<\epsilon$ whenever $|x-2k\pi|<\delta_k$, and $|f(x)-3|<\epsilon$ whenever $\left|x-\frac{(4k+1)\pi}{2}\right|<\delta_k'$. Now by this problem we can show that $2k\pi$ and $\frac{(4k+1)\pi}{2}$ is arbitrarily close to some integer as $k$ varies over $\mathbb{Z}$. Hence $f(n)$ oscillates between $2$ and $3$ for $n\in\mathbb{N}$.