So, as is in the title, I'm required to prove that $$\lim_{x \to 0^+} \sin(\frac{1}{x})$$
does not exist.
Proof Attempt: Suppose for a contradiction that $$\lim_{x \to 0^+} \sin(\frac{1}{x}) = l$$
for some $l \in [-1, 1]$.
Then, by the sequential criterion for right hand limits, we may conclude that for every sequence $\{a_j\}$ for which $a_j \in (-\infty, 0)\cup(0, \infty)$ and $a_j > 0 \quad \forall j \in \mathbb{N}$ that converges to $0$, $\sin(a_j)$ converges to $l$.
Consider the sequence $$b_j = \frac{2}{(4j+1)\pi}$$
We observe that clearly this sequence converges to $0$ and $b_j > 0$ for $j = 1, 2, \ldots$
Thus, $$\lim_{j \to \infty} \sin(\frac{1}{b_j}) = l$$
But we see that $$\sin(\frac{1}{b_j}) = \sin(\frac{(4j+1)\pi}{2}) = 1$$
Hence,
$$\lim_{j \to \infty} \sin(\frac{1}{b_j}) = \lim_{j \to \infty} 1 = 1$$
So, we may conclude that $l = 1$. However, now consider the sequence $\{c_j\}$ given by $c_j = \frac{1}{j\pi}$. Clearly $c_j \to 0$ and $ c_j > 0$ for $j = 1, 2, \ldots$
Therefore, by the sequential criterion for right hand limits, we must have:
$$\lim_{j \to \infty} \sin(\frac{1}{c_j}) = l = 1$$
But we see that $\sin(\frac{1}{c_j}) = \sin(j\pi) = 0$ for $j = 1, 2, \ldots$
Therefore, $$\lim_{j \to \infty} \sin(\frac{1}{c_j}) = \lim_{j \to \infty} 0 = 0 \neq 1$$
This is a contradiction. Therefore, $$\lim_{x \to 0^+} \sin(\frac{1}{x}) \quad \text{DNE}$$
Is my proof correct? Are there any suggestions/improvements that I may make?
Best Answer
Looks OK except some typos: you should have $\sin \frac{1}{b_j}$ instead of $\sin b_j$.
You don't need to prove the statement by contradiction and the proof could be written in a concise way:
Define two positive sequences $(a_n)$ and $(b_n)$ such that $$ \frac{1}{a_n}=n\pi,\quad \frac{1}{b_n}=2n\pi+\frac{\pi}{2}. $$ Note that $a_n\to 0$ and $b_n\to 0$ as $n\to\infty$. But we have $$ \lim_{n\to\infty}\sin\left(\frac{1}{a_n}\right)=0,\quad \lim_{n\to\infty}\sin\left(\frac{1}{b_n}\right)=1. $$ Hence by the sequential criterion, the limit $\lim_{x\to 0+}\sin \frac{1}{x}$ does not exist.