- #1
Usagi
- 45
- 0
Show that $\lim_{x \rightarrow 0} \frac{1}{x}$ does not exist.
Is my following argument correct?
I will show there exists a sequence $(x_n) \subset \mathbb{R} \backslash \{0\}$ satisfying $x_n \neq 0$ and $(x_n) \rightarrow 0$, but $\lim_{n \rightarrow \infty} f(x_n)$ does not exist. Consider $(x_n) = \frac{1}{n}$ for $n \in \mathbb{N}$, clearly $x_n \neq 0$ and $(x_n) \rightarrow 0$, but $f(x_n) = n$, which diverges to $+\infty$ as $n \rightarrow \infty$, hence $\lim_{x \rightarrow 0} \frac{1}{x}$ does not exist.
Is my following argument correct?
I will show there exists a sequence $(x_n) \subset \mathbb{R} \backslash \{0\}$ satisfying $x_n \neq 0$ and $(x_n) \rightarrow 0$, but $\lim_{n \rightarrow \infty} f(x_n)$ does not exist. Consider $(x_n) = \frac{1}{n}$ for $n \in \mathbb{N}$, clearly $x_n \neq 0$ and $(x_n) \rightarrow 0$, but $f(x_n) = n$, which diverges to $+\infty$ as $n \rightarrow \infty$, hence $\lim_{x \rightarrow 0} \frac{1}{x}$ does not exist.