Why can we simplify $(x^2-4)/(x-2)$ to get $x+2,$ but what happens at $x=2$ when $y=(x^2-4)/(x-2)?$ When teaching algebra I have always taught that we can simplify $\dfrac{x^2-4}{x-2}$ by factorising the top line and "cancelling" the $(x-2)$ in the top and bottom because they are a factor of $1.$
But what about if $x=2\,?$ I have recently been sketching functions and was asked to sketch $\dfrac{f(x)}{g(x)}$ where $f(x)=x^2-4$ and $g(x)=x-2$ and the answer in the book has the line $y=x+2$ with the point where $x=2$ removed.
Are these questions subtly different or is the simplification technically incorrect? Does it have an implication that $x$ is not equal to $2?$
 A: All we are doing is: the following equation holds for every $x \in \mathbb{R}-\{2\}$:
\begin{equation*}
\frac{x^2-4}{x-2}=x+2
\end{equation*}
We exclude the case $x=2$ because the left side is undefined when $x=2$ so that, when thinking the left side as a function we exclude $2$ from its domain.
More details:
When we define a function, three things must be specified: the domain, the codomain, and the rule which the function uses to associate each element in the domain an element in the codomain.
If $F$ and $G$ are functions, we write $F=G$ to express the meaning "$F$ and $G$ are the same function". That is:

*

*$F$ and $G$ have the same domain.

*$F$ and $G$ have the same codomain.

*For each $x$ belonging to the domain, $F(x)=G(x)$.

When people write an expression in format $f(x)=$ something, where something involves a polynomial, a quotient, an exponential, or any other things that makes sense when $x$ is a real number, without specifying anything, we tacitly assume:

*

*$f$ is a function.

*The domain of $f$ is whatever subset of the real line such that the expression "something" is well-defined.

*The codomain of $f$ is the real line.

*If $x$ belongs to the domain, then $f(x)$ is given by this "something".

Now, as an example, we go back to your question. When we write $f(x)=x^2-4$, what we mean, tacitly, is that we define a function $f:\mathbb{R} \to \mathbb{R}$ by setting $f(x)=x^2-4$ for every $x \in \mathbb{R}$ simply because the expression makes sense for every $x \in \mathbb{R}$. Similarly, when we write $g(x)=x-2$, we mean that we define a function $g:\mathbb{R} \to \mathbb{R}$ by setting $g(x)=x-2$ for every $x \in \mathbb{R}$. By writing $y=\frac{f(x)}{g(x)}=\frac{x^2-4}{x-2}$, we mean that we define a function $y: \mathbb{R}-\{2\}\to \mathbb{R}$ by setting $y(x)=\frac{x^2-4}{x-2}$ for every $x \neq 2$, simply because the expression makes sense for every $x \neq 2$.
Now, if we define a function $z: \mathbb{R} \to \mathbb{R}$ by setting $z(x)=x+2$, then we may write $z \neq y$ because their domain is different.
Although one may remark that $z(x)=y(x)$ for every $x \neq 2$, this is irrelevant to our current dicussion about equality between functions. (But as pointed out in the comment, since the domain of $y$ is a subset of the domain of $z$, sometimes people still write $z=y$ to express the above remark. This is not standard.)
Finally, by the graph of a function $F:A \to B$, we mean the set $\{(a,F(a))\}_{a\in A}\subseteq A\times B$. If we use $Y$ to denote the graph of $y$ and $Z$ for graph of $z$, then the remark shows that $Y$ and $Z$ differ by exactly one point, namely the point $(2,4) \in \mathbb{R}^2$.
A: 
we can simplify $\dfrac{x^2-4}{x-2}$ by "cancelling" the $(x-2)$
Does it have an implication that $x$ is not equal to $2?$

Indeed, this simplification is valid only if $x$ doesn't equal $2:$
\begin{align}\frac{(x+2)(x-2)}{x-2}=x+2&\implies x\ne2.\end{align}

is the simplification technically incorrect?

It depends on the context. While $$\frac{(x+2)(x-2)}{x-2}\not\equiv x+2,$$ the simplification is correct when, for example, we are dealing with only negative values: \begin{align}x<0 &\implies \frac{(x+2)(x-2)}{x-2}=x+2.\end{align}
In general, $$y=\frac{h(x) \times g(x)}{g(x)} \\\iff \\y=h(x)\quad\text{and}\quad g(x) \ne0.$$
