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Question

Let's say that we have the following

$f(x)=\frac{x+1}{x^2-1}$

The next thing that we would do is to factor the denominator to get the following:

$f(x)=\frac{x+1}{(x+1)(x-1)}$ then we next divide both the numerator and the denominator by (x+1) to get:

$\frac{1}{x-1}=f(x)$

But then If we look at the equation we would presume that there is an asymptote which is x=1. However if we plug in x=-1 why is that undefined and not an asymptote?

To clarify what I ask for is how to know when it is an asymptote and when it is just a point that is undefined algebraically. (sure we can use a graph but that is tedious)

1 Answers 1

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It is called a removable singularity. In essence, a point is called a vertical asymptote if

$$\lim_{x\to a}f(x)=\pm\infty$$

In words: as $x$ gets very close to $a$, $f(x)$ gets closer to $\pm\infty$.

Clearly, this will be the case as $x\to1$, since we'll get division by $0$, or rather, division by a very small number.

On the other hand, as $x\to-1$, then $f(x)\to-\frac12\ne\pm\infty$, so it is not an asymptote, but a removable singularity.

To conclude, it is not undefineability that determines asymptotes, but how the function behaves.

Also, you can't plug in $-1$, since you'll get $f(x)=\frac0{-2\times0}=\text{undefined}$. The algebraic cancellations you did only work if $x\ne-1$.

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    One might argue that the limit relies on knowledge of the graph of a function in principle. I personally consider it a property of the geometric interpretation. Great answer though! – 2017-01-21