<s>[INST] 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 $2?$
 [/INST] 
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, $$y=\frac{h(x) \times g(x)}{g(x)} \\\iff \\y=h(x)\quad\text{and}\quad g(x) \ne0.$$
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