Vertical asymptote, yes or no?

I am working on a problem that will highlight the importance of accuracy and the flaw in approximating certain numbers (very basic stuff).

Say you have the following function

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

If I were to draw this graph, I would get a straight line with a "gap" at $x = -b$.

But if I were to approximate $b$ in the following manner

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

where $\tilde{b}$ is equal to some sort of approximation of $b$ (so imagine $b=\pi$ and $\tilde{b}=3.1415$).

Now, if I were to draw this graph, the result would be a straight line like the one I previously drew, but the closer I got to $-\tilde{b}$, the graph would (depending on which direction I came from) jump in two directions.

I no longer have that "unspottable gap" that I had previously. Why not? Why does it now jump in two directions instead of just behaving like a straight line (but with that "gap" at $-\tilde{b}$ still being there, just not visible)?

And which one of these would have "a vertical asympote" and which one would not?

• I LaTeX-ed your question. It's useful to know How to write math. Doeas this edit reflect what you intended to ask? – cjferes Aug 26 '14 at 22:26

If I were to draw this graph, I would get a straight line with a "gap" at $x = -b$.

But if I were to approximate b in the following manner

$$\frac{x^2 - b^2 }{ x + \bar{b}}$$

where $b$ is equal to some sort of approximation of $b$ (so imagine $b$ is $\pi$ and $\bar b$ is 3.1415).... now, if I were to draw this graph, the result would be a straight line like the one I previously drew, but the closer I got negative 3.1415 (in other words, negative $b$), the graph would (depending on which direction I came from) jump in two directions.

Right, so, the main case in dealing with this discontinuity lies with the fact that there is a 'removable' discontinuity at $x=-b$ (where the limit as $x\to -b$ exists but $f(-b)$ is not defined): $$\frac{x^2-b^2}{x+b}=\frac{(x-b)(x+b)}{x+b}=x-b;\;x\ne -b$$But, if we take some approximation $\bar b=b+\epsilon$, for some $\epsilon\ne 0$, then we run into some problems because the fraction does not cancel as: $$\frac{(x-b)(x+b)}{x+\bar b}=\frac{(x-b)(x+b)}{x+b+\epsilon}$$ If we shift the graph by allowing $\gamma=x+b+\epsilon$, then: $$\frac{(x-b)(x+b)}{x+\bar b}=\frac{(\gamma-2b-\epsilon)(\gamma-\epsilon)}{\gamma}$$ As we take the limit as $\gamma(x)=x+b+\epsilon\to 0$ (which is what you did when finding the asymptote) we get, for small $\gamma$: $$\frac{(\gamma-2b-\epsilon)(\gamma-\epsilon)}{\gamma}\approx\frac{C\cdot \epsilon}{\gamma}\to \pm\infty, \;\gamma\to 0^\mp.$$For some constant $C<0$.

This is a good example to show that some discontinuities are numerically unstable on most floating point schemes.

So, to be more specific and directly answer your questions, the gap that you now had does not exist as you have some residue term as $x\to -b$. Of course, no matter how small the term $\epsilon$, if you have $\gamma\to 0^\pm$ with constant $\epsilon$, at some point $|\epsilon/\gamma| > M$ for any $M$. We can fix this by assuring that the residual term grows small as the limit term is in some neighbourhood of its limit, but that requires some set of assumptions on the approximation or on the floating-point scheme.

The vertical asymptote, on the other hand, happens only when the terms do not cancel.

EDIT: Added some more stuff and fixed some statements for clarity.