Questions
Let $D=\{\frac{1}{n}:n\in\mathbb N\}\subseteq\mathbb R$.
(a) Show that $0$ is a cluster point of D
(b) Let $f: x \mapsto1$ where $x\in D$. Show that $\lim_{x\to\infty} f(x) = 1$.
(c) Let $g: \frac{1}{n} \mapsto k$ where $\frac{1}{n}\in D$ and $k\in\mathbb N_0$ is the largest number such that $2^k | n$. Show that $\lim_{x\to\infty}g(x)$ does not exist
My answers
(a)
Theorem: Let $D\subseteq\mathbb R$ and let $x\in\mathbb R$. $x$ is a cluster point if and only if $\exists (x_k)_{k\in\mathbb N}$ such that $x_k\in D\setminus{x}$ for all $k\in\mathbb N$ and $\lim_{k\to\infty}x_k=x$.
Let $x_n=\frac{1}{n}$ where $n\in\mathbb N$ be a sequence. We can define D in terms of $x_n$.
i.e $D=\{x_n\}$. So $x_n\in D$ $\space\forall n\in\mathbb N$.
Proving that $0\notin D$: Assume $0\in D$ $\Leftrightarrow \space \exists n\in\mathbb N$ such that $\frac{1}{n}=0 \space$ $\Leftrightarrow 1=0$ $\therefore$ Contradiction $\Rightarrow 0\notin D$ $\Rightarrow D\setminus\{0\}=D$
Therefore $x_n\in D\setminus\{0\}$
$\forall \epsilon\gt 0\space$ $\space\exists N_{\epsilon}\in\mathbb N$ where $N_{\epsilon}=\frac{1}{\epsilon}$ : $\forall n\gt N_{\epsilon}\space$ $\Rightarrow |x_n-0|=|\frac{1}{n}-0|=\frac{1}{n}\lt\frac{1}{N_{\epsilon}}=\epsilon$
Hence $\lim_{n\to\infty}x_n=0$
$\Rightarrow$ By the theorem provided we have shown that $x=0$ is a cluster point.
(b)
Given $\epsilon\gt0$. We can choose any $\delta$ with $0\lt |x-0|\lt\delta$ we get $|f(x)-1|=|1-1|=0\lt\epsilon$
Hence $\lim_{x\to 0}f(x)=1$
(c) I have tried for a while now -struggling to see how to tackle this.
Comments
This is a nasty question which was posed to me on my Analysis I course. Would be great if anyone could check my work, maybe even give alternative proofs for (a) and (b). And it would be nice if someone could help me with (c) :)
(a) and (b) are fine. For (c), you have for example that $$g\left(\frac{1}{ 3^k}\right)=0\quad \text{and}\quad g\left(\frac{1}{2\cdot 3^k}\right)=1,$$ for all $k\in\mathbb N$. Therefore, if $$x_k=\frac{1}{3^k }\quad \text{and}\quad y_k=\frac{1}{2\cdot 3^k},$$ then $$\lim_{k\to \infty } g(x_k)=0<1=\lim_{k\to \infty }g(y_k),$$ and thus, $g$ has no limit in $+\infty $.