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When I was playing with numbers, I found that it appears to be true that only if given the fact that some arbitrary function has asymptotes at any $x=\pm n_{a+a\,\ldots}$ it satisfies the function

$$g=\frac{1}{(x\pm n_{a+a_0})(x\pm n_{a+a_1})\ldots}$$

This fact can also be used when you want to find the asymptotes of a function; take for example the function

$$y=\frac{x^2-1}{x^2-4}$$ Simplify the denominator, $$\frac{x^2-1}{(x-2)(x+2)}$$ As we in the denominator have that $(x-2)$ and $(x+2)$ there exists asymptotes at $x=2$ and $x=-2$ respectively.

If assuming that this is true, why is this true? I understand that there exists a correlation between $x=\pm n_{a+a_0}$ and $(x\pm n_{a+a_0})$.

Left as a comment:

It may only be very simple explanation(s) to what I have "found", but if anyone have the time I would like to hear the explanation(s) anyways for clarification. I have only just recently learned what asymptotes is, and that only in a very intuitive way (they didn't even speak about limits). Sorry for eventual "unmathematicalness" in my reasoning further up.

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For any rational function $f(x)=\frac{g(x)}{h(x)}$, there are vertical asymptotes at the zeros of $h$, which are the points $x$ such that $h(x)=0$. This is because $f$ is undefined at these points.

It is not true that every function with vertical asymptotes can be written in the form you describe; consider $f(x)=\tan x$.

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