Using sequential definition of functional limits, show that $\lim_{x \rightarrow 0} 1/x$ does not exist
I have two questions regarding this. Firstly, say we have a function that 'converges' to $+\infty$, e.g., $\lim_{x \rightarrow 0}1/x^2 = +\infty$, do we say this limit "exists"? Or do limits only exist for finite values? I.e., if the limit of a function is $\pm \infty$, then is it correct to say that the limit doesn't exist?
Second, is my following argument correct?
Define $f: \mathbb{R} \backslash \{0\} \rightarrow \mathbb{R}, f(x) = 1/x$. I claim that $\lim_{x \rightarrow 0} \frac{1}{x}$ does not exist. I will use the divergence criterion for functional limits to show this. Define $(x_n) = 1/n$ and $(y_n) = -1/n$, $n \in \mathbb{N}$ which are both sequences in $\mathbb{R} \backslash \{0\}$. Note that $x_n \neq 0$ and $y_n \neq 0$ for all $n$. Furthermore, $f(x_n) = n$ and $f(y_n) = – n$. Observe that $\lim x_n = \lim y_n = 0$, but $\lim f(x_n) = + \infty$ and $\lim f(y_n) = – \infty$, hence $\lim f(x_n) \neq \lim f(y_n)$, thus $\lim_{x \rightarrow 0} \frac{1}{x}$ does not exist.
Best Answer
Your proof is entirely correct.
The question about whether or not '$f(x_n) \rightarrow \infty$' is said to converge is a good one. Technically, the limit does not exist since the 'limiting value' is not an element of the real numbers. In an undergraduate course, you can say that the limit 'diverges to $\infty$', or simply that the limit is $\infty$ and therefore does not exist.
On the other hand, it is often useful to think of $\infty$ as being a well-defined limit point, and so people often consider the extended real numbers $\overline{\mathbb{R}}:=\mathbb{R}\cup\{\pm \infty\}$. We then say that $y_n \rightarrow \infty \in \overline{\mathbb{R}}$ if for any $\alpha > 0$, there is $N \in \mathbb{N}$ such that for all $n \geq N$, $y_n > \alpha$.