This is a real nightmare, for university teachers. Every student of mine comes to my class from high school and is sure that $x \mapsto 1/x$ is discontinuous at $0$. The reason why calculus textbooks are so ambiguous is that their authors do not like to leave something undiscussed. For some reason, the answer to any question should be "yes" or "no"; hence they tend to formally state that functions are discontinuous outside their domain of definition.
In my opinion, this is a very bad approach: it is a matter of fact that discontinuous should not be read as the negative of continuous. The domain of definition makes a difference, and the most useful idea is that of continuous extension.
Almost any mathematician would say that the tangent function is continuous inside its own domain of definition.
Two functions are typically defined to be equal if and only if they...
- Share the same domain
- Share the same codomain
- Take on the same values for each input.
Thus, functions $f,g : S \to T$ for sets $S,T$ have $f=g$ if and only if $f(x) = g(x)$ for all $x$ in $S$.
For functions with holes, we typically restrict the domain by ensuring the values where the function is not defined at not included. For example, in the functions you have, you have
$$f(x) = \frac{(x-3)(x+2)}{(x-3)} \;\;\;\;\; g(x) = x+2$$
Are these equal? Yes, and no.
A function must be defined at all values of the domain. Thus, we can say $3$ is not in the domain of $f$ for sure. But we never specified otherwise the domains and codomains of these functions! Typically, unless stated otherwise, we often assume their domain to be $\Bbb R$ or $\Bbb C$, minus whatever points are causing problems - and of course, in such cases, $f \neq g$ since $f(3)$ is not defined, and thus $f$ normally has domain $\Bbb R \setminus \{3\}$ and $g$ generally has domain $\Bbb R$.
But that restriction is not necessary. For example, we could define the functions to be $f,g : \Bbb R \setminus \Bbb Q \to \Bbb R$. Notice that the domain of both functions are now all real numbers except rational numbers, i.e. the irrational numbers. This means $3$ is not in the domain of either function - and since that's the only "trouble spot," and the codomains are equal, and the values are equal at each point in the domain, $f=g$ here.
Or even more simply: we could have $\Bbb R \setminus \{3\}$ be the domain of $f$ and $g$ and again have equality! The key point in all this is that, just because $f$ or $g$ do attain defined values for certain inputs, doesn't mean they have to be in the domain.
In short, whether $f=g$ depends on your definitions of each. Under typical assumptions, $f \neq g$ in this case, but if we deviate from those assumptions even a little we don't necessarily have inequality.
Best Answer
This function is continuous everywhere in its domain of definition.
Its domain of definition is $\mathbb R\setminus \{-1,0,1\}$. That's because the expression $1/\log(|x|)$ only makes sense on that set, since the expressions $1/\log(|-1|)$, $1/\log(|0|)$, and $1/\log(|1|)$ are meaningless. This does not make these points of discontinuity, it simply means they are not in the domain of the function -- a completely separate concept.
This is not to say, however, that the function doesn't have a continuous extension to include $\{0\}$ in the domain, but that's not the same question.