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.
The map$$\begin{array}{rccc}f\colon&\mathbb R\setminus\{0\}&\longrightarrow&\mathbb R\\&x&\mapsto&\frac{x^2}x\end{array}$$is undefined at $0$, and therefore it is meaningless to ask whether or not it is differentiable there. It happens that we can extended it to one and only one continuous function $F\colon\mathbb R\longrightarrow\mathbb R$, which is defined by $F(x)=x$. And it happens that this function is differentiable at $0$.
However, if you take$$\begin{array}{rccc}g\colon&\mathbb R\setminus\{0\}&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}x&\text{ if }x>0\\-x&\text{ if }x<0,\end{cases}\end{array}$$then you can extend $g$ to one and only one continuous map $G\colon\mathbb R\longrightarrow\mathbb R$, which is $G(x)=\lvert x\rvert$, but the function $G$ is not differentiable at $0$.
Best Answer
Even though it is written as a condition, $f(x)$ is in fact a function.
To find a removable discontinuity, we want to have a point such that $$\lim_{x\rightarrow x_0+} f(x) =\lim_{x\rightarrow x_0-} f(x) \neq f(x_0)$$ where the limits are those from the left and the right.
Note that at $x=0$, we have that the limit approaching $0$ from the right of $f$ and the limit approaching $0$ from the left of $f$ are both $0$ but $f(0)=1$. Therefore there is a removable discontinuity at $x = 0$.