Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm currently learning about asymptotes and I'm trying to work out the following example:

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

To find the vertical asymptotes, the book I'm following says that after factoring completely, you should set each factor of the denominator to $0$ and:

Every solution you get that does not make the numerator 0 will give you a vertical asymptote of the function.

According to that, I do:

$$ \begin{align} x^2+1=0 \\ x^2=-1 \\ x = \pm\sqrt{-1} \\ x = \pm i \end{align} $$

Now, neither $i$ nor $-i$ make the numerator $0$, so does that mean that $x=i$ and $x=-i$ are vertical asymptotes to the function $f$? And if so, what does that really mean?

Wolfram|Alpha doesn't mention any vertical asymptotes, only the horizontal one at $y=0$.

share|improve this question
5  
Do note that $x^2+1$ is never zero when $x$ is real. Complex asymptotes are usually called poles in this context (of a ratio between two polynomials). –  Asaf Karagila Oct 9 '11 at 21:54

4 Answers 4

up vote 7 down vote accepted

To illustrate what 3296 was getting at, consider these two plots:

real and imaginary parts of 1/(z^2+1)

These are plots of the real and imaginary parts of the function $\dfrac1{z^2+1}$, where $z=x+i y$. The poles of the function at $z=\pm i$ are easily seen in the plots. Since we're limited to seeing (a two-dimensional projection of) three dimensions, we are forced here to illustrate the poles by plotting the real and imaginary parts of the function separately.

Now, look at a "slice" of the real part:

slice of the real part

You see at once the plot of the function you are accustomed to on the real line. The poles do not lie in the slice, and this corresponds to you seeing no vertical asymptotes in the plots of your function on the real line.

Incidentally, this function is the usual example for demonstrating the so-called "Runge phenomenon": any attempt to approximate this function with a polynomial fails due to the poles in the complex plane, even if you are only considering real values of the argument.

share|improve this answer

Let me attempt to answer this in a more direct way that avoids allusions to complex analysis "in the future."

When you graph this function on a 2D piece of paper, you are graphing a real input versus a real output. You get a one-dimensional curve in two-dimensional space. An asymptotes is a meaningful property of a one-dimensional curve embedded in a larger space.

If you were to graph this function as a complex function, regarding the complex numbers as living on the (two-dimensional) complex plane, you'd have two input axes and two output axes, for a total of four dimensions. The graph of the function would be a two-dimensional surface in four-dimensional space. Obviously this situation is going to be substantially more complicated, so it's a safe assumption that you're only going to be asked to deal with the simpler one-dimensional problem above.

share|improve this answer

Since presumably your domain is the real numbers, the fact that $\pm i$ causes the denominator to be 0 is of no consequence to you. That is, if you run over all the real numbers and plug them in one-by-one to the function, you're never going to see $x = i$ or $x = -i$ because they are not real numbers. Since you're never going to select those two "bad" $x$'s, you're never going to divide by 0, and so you're never going to have a place where asymptotes might arise.

share|improve this answer

The "asymptote" concept is mostly used only in real analysis, so when one asks about asymptotes it's generally implied that we're considering the function only on the real axis.

In the complex case we'd say that the function has poles at $i$ and $-i$, which corresponds more or less to vertical asymptotes (but in fact have much nicer and subtler properties). You shouldn't need to be concerned about such things until you get to complex analysis.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.