Find all the limit points of the set $\{(x,y); y=2\cos(1/x)+1,x>0\}$

Question

Find all the limit points of the set $$\{(x,y); y=2\cos(1/x)+1,x>0\},$$ which do not lie on the curve $$y=2\cos(1/x)+1.$$

Answer 1

Let $(a,b)$ be a limit point which does not lie on the curve. Then there is sequence $(x_n,y_n)$ such that $y_n=2 \cos(\frac 1 {x_n})+1$ and $(x_n,y_n) \to (a,b)$. If $a \neq 0$ then we get $b=2cos (\frac 1 a)+1$, but this is not permissible. So $a=0$. It is clear that $-1 \leq y_n \leq 3$ for all $n$ so $-1 \leq b \leq 3$. Now we prove that any point $(a,b)$ with $a=0$ and $-1 \leq b \leq 3$ is a limit point (which of course does not lie on the curve). For this what you need is the following: given any $c \in [-1,1]$ there is a sequence $(x_n)$ such that $x_n \to 0$ and $\cos (x_n) \to c$. This is easy: just take $x_n=\frac 1 {{2n\pi}+d}$ where $d=\cos^{-1} c$.

Answer 2

The range of $f(x)=2\cos{1\over x}+1$ is $[-1,3)$, therefore $$[-1,3)\subseteq L\subseteq[-1,3]$$where $L$ is the set of limit points. Also $3$ is achievable with the sequence $x_n=n$, hence$$L=[-1,3]$$