Is $ f(x) = {x \over x} $ well-defined at $x = 0$?

Question

Suppose that f is a function which domain and range are real numbers.

Then, If $ f(x) = {x \over x} $ then Is f well-defined at $x = 0$?

In other words, does ${x \over x} = 1$ equal on $ \mathbb{R}$?

Answer 1

No, because $f(0)=\frac{0}{0}$. However, because this function has a limit at $0$ (and everywhere else), you can make a continuous, well-defined function from it given by $\tilde{f}(x)=\lim\limits_{y\to x}f(y)$.

Answer 2

Not. However, you have some options:

1. You can restrict the domain to $ \mathbb{R} - \{0\}$.

2. You can redefine a function $\tilde {f}$ by

$$ \tilde {f}(x)= \begin{cases} f(x) & \text{if} & x\neq0,\\ 1 & \text{if} & x=0.\\ \end{cases} $$