Reciprocal Function is Unbounded

[Bashyboy]

Homework Statement

Let $f : (0,\infty) \to \Bbb{R} - \{0\}$. If $\frac{1}{f}$ is unbounded on every interval containing $x=3$, will $\lim_{x \to 3} f(x) = 0$?

Homework Equations

The Attempt at a Solution

Consider the function

$$f(x) = \begin{cases} 1, & x \in \Bbb{Q} \cap (0,\infty) \\ x-3, & x \notin \Bbb{Q} \cap (0, \infty) \\ \end{cases}$$

Let $x_n \notin \Bbb{Q} \cap (0,\infty)$ be an irrational sequence converging to $3$. Then $\lim_{n \to \infty} \frac{1}{f(x_n)} = \lim_{n \to 3} \frac{1}{x_n -3} = \pm \infty$ (I am abusing notation out of laziness). This shows that $\frac{1}{f}$ is unbounded on every interval of containing $3$. Now let $r_n \in \Bbb{Q} \cap (0, \infty)$ be a rational sequence converging to $3$. Then $\lim_{n \to \infty} f(r_n) = \lim_{n \to \infty} 1 = 1 \neq 0$.

How does this sound? What if $f$ is required to be continuous; will the theorem hold?

 

[andrewkirk]

Yes that is correct. If $f$ is continuous the theorem will not hold - although that needs to be proven.