Below the following question is my solution but I am not sure whether the proofs are correct and rigorously exhaustive so, if present, kindly point out any errors.
Prove or disprove (by providing a counterexample) each of the following statements:
$(i)$ Let $\{x_n\}$ be a sequence such that $\underset{\rm n}{\rm lim}$ $x_n = 0$, then $\underset{\rm n}{\rm lim}$ $ \frac{1}{|x_n|} = +\infty$
$(ii)$ Let $\{y_n\}$ be a sequence such that $\underset{\rm n}{\rm lim}$ $y_n = 0$, then either $\underset{\rm n}{\rm lim}$ $ \frac{1}{y_n} = +\infty$ or $\underset{\rm n}{\rm lim}$ $ \frac{1}{y_n} = -\infty$
Answer:
$(i)$ Let $M \in \mathbb{R}, M>0$, we need to show that $N \in \mathbb{N}$ such that $\frac{1}{|x_n|}>M, \hspace{5pt} \forall n \ge N$. As $\underset{\rm n}{\rm lim}$ $x_n = 0$ for $\zeta = \frac{1}{n}>0, \hspace{5pt} \exists n \in \mathbb{N}$ such that
$|x_n|<\frac{1}{M} \hspace{5pt} \forall n \ge N$ $\Rightarrow$ $\frac{1}{|x_n|}>M \hspace{5pt} \forall n \ge N$ $\Rightarrow$ $\underset{\rm n}{\rm lim}$ $ \frac{1}{|x_n|} = +\infty$. (Hence statement $(i)$ is true).
$(ii)$ Using a counterexample to disprove the statement, let $y_n = \frac{(-1)^n}{n}.$ Then $\underset{n\rightarrow \infty}{\rm lim} y_n =0. $ And $\underset{n\rightarrow \infty}{\rm lim} y_n$ is neither $+ \infty$ nor $- \infty$ (since $\underset{n\rightarrow \infty}{\rm lim} y_n$ does not exist). Here $y_n$ is an oscillating sequence. Therefore, the statement is false.
Best Answer
Outside of a minor typo, your analysis for part (i) is valid (and accurate). Typo :
$\exists n \in \mathbb{N}$ such that...
should be
$\exists N \in \mathbb{N}$ such that
Your conclusion for part (ii) is wrong and your counter example is invalid.
In part (ii), the assertion concerns $\frac{1}{|y_n|}$ while your counter example is incorrectly focusing on $y_n$.
More to the point, if you inspect the assertions of part (i) and (ii) closely, you will see that part (ii) requires no work.
This is because, by the wording of parts (i) and (ii), it is impossible for part (ii) to be false if part (i) is true.
Further, you have already (validly) proven that part (i) is true.
Therefore, part (ii) must be true.
As an irrelevancy, looking more closely at the trailing possibility in part (ii), you will see that it is nonsensical because both the numerator and denominator must be non-negative. Therefore, how can the fraction be going to $-\infty?$
Addendum
Reaction to OP's editing of the query to remove the absolute value sign in the denominators, re part (ii).
Yes, this editing permits part (ii) to be false, re $y_n$ can oscillate between positive and negative.