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I am asked to find the vertical and horizontal asymptotes of the equation:

$$f(x)=(a^{-1}+x^{-1})^{-1}$$

I simplify this to

$$f(x)=\frac{1}{a^{-1}+x^{-1}}$$ $$f(x)=a^1+x^1$$$$f(x)=a+x$$Which is some constant, graphed as horizontal line - that will not have a vertical or horizontal asymptote. Is my algebra terribly off?

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    $\begingroup$ I believe the common error $$(a+b)^n=a^n+b^n$$ is known as the "Freshman's dream" :). See en.wikipedia.org/wiki/Freshman%27s_dream $\endgroup$
    – Argon
    Jun 18, 2012 at 1:39
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    $\begingroup$ @Argon I was thrown into a fit of uncontrollable laughter the first time I learned the formula is actually correct in a field of prime characteristic $n$. $\endgroup$
    – Austin Mohr
    Jun 18, 2012 at 2:07

3 Answers 3

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If $\displaystyle f(x) = \frac{1}{a^{-1} + x^{-1}}$, we can't distribute the inversion $x \mapsto x^{-1}$ to say that $f(x) = a + x$.

A concrete example: $\displaystyle \frac{1}{\frac{1}{2} + \frac{1}{3}} = \frac{1}{\frac{5}{6}} = \frac{6}{5} \neq \frac{1}{\frac{1}{2}} + \frac{1}{\frac{1}{3}} = 2 + 3 = 5$.

(We can, however, distribute the divisor like this: $\displaystyle \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$.)

Note that the function is defined in $\mathbb{R} \setminus\{-a\}$ (the only "problem" we might run into in order to evaluate $f(x)$ at an arbitrary $x \in \mathbb{R}$ is dividing by zero; and that would only happen when $a^{-1} + x^{-1} = 0 \Leftrightarrow -a = x$), so to check if there are vertical asymptotes, we calculate $\lim_{x \to -a} f(x)$.

$\displaystyle \lim_{x \searrow -a} \frac{1}{a^{-1}+x^{-1}} = \lim_{x \nearrow -a^{-1}} \frac{1}{a^{-1} + x} = \lim_{x \nearrow 0} \frac{1}{x} = - \infty$, and likewise

$\displaystyle \lim_{x \nearrow -a} \frac{1}{a^{-1}+x^{-1}} = \lim_{x \searrow -a^{-1}} \frac{1}{a^{-1} + x} = \lim_{x \searrow 0} \frac{1}{x} = + \infty$.

(Note that $x \geq -a \Rightarrow x^{-1} \leq (-a)^{-1}$ and $x \leq -a \Rightarrow x^{-1} \geq (-a)^{-1}$ if $x$ and $-a$ have the same sign, which will happen when they are close enough, for example, in this case because we're taking a limit; that's why when we start approaching a limit from the left and make the substitution we end up approaching the limit from the right and vice versa.)

We deduce that there is a vertical asymptote with equation $x=-a$.

As for the horizontal asymptotes, we calculate $\lim_{x \to \pm \infty} f(x)$, and we'll get:

$\displaystyle \lim_{x \to \pm \infty} \frac{1}{a^{-1} + x^{-1}}= \lim_{x \to 0} \frac{1}{a^{-1}+x} = \frac{1}{a^{-1}} = a$, so there is a horizontal asymptote with equation $y=a$.

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    $\begingroup$ +1 for giving a concrete example demonstrating the incorrect algebra. $\endgroup$
    – Austin Mohr
    Jun 18, 2012 at 2:08
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\begin{align} f(x) & = (a^{-1} + x^{-1})^{-1} = \dfrac1{a^{-1} + x^{-1}}\\ & = \dfrac1{\dfrac1a + \dfrac1x} = \dfrac1{\dfrac{x+a}{ax}}\\ & = \dfrac{ax}{x+a} = \dfrac{ax + a^2 - a^2}{x+a} \text{(Adding and subtracting $a^2$ to the numerator)}\\ & = \dfrac{ax+a^2}{x+a} - \dfrac{a^2}{x+a} = \dfrac{a(x + a)}{x+a} - \dfrac{a^2}{x+a}\\ & = a - \dfrac{a^2}{x+a} \end{align} Now note that as $x \to -a^{-}$, we have $f(x) \to + \infty$ and as $x \to -a^+$, we have $f(x) \to -\infty$.

Hence, $x=-a$ is a vertical asymptote.

Now letting $x \to +\infty$, we get that $f(x) \to a^{-}$ and $x \to -\infty$, we have that $f(x) \to a^+$.

Hence, $f(x) = a$ is a horizontal asymptote.

Below is a plot of this curve in blue with $a=2$. The red line indicates the horizontal asymptote i.e. $f(x) = a$. The pink line indicates the vertical asymptote i.e. $x=-a$.

enter image description here

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  • $\begingroup$ I am a little mixed up on the final function you arrived at, as it differs from the function arrived at below. Is it just that you used the distributive property to move the $a$ out in front of the function, if so, is there a reason for this in illustrating your example? $\endgroup$
    – Kurt
    Jun 18, 2012 at 2:09
  • $\begingroup$ @Kurt The function is $\dfrac{ax}{x+a}$. Now add and subtract $a^2$ from the numerator. Hence, $$\dfrac{ax}{x+a} = \dfrac{ax+a^2 - a^2}{x+a} = \dfrac{ax+a^2}{x+a} - \dfrac{a^2}{x+a} = \dfrac{a(x+a)}{x+a} - \dfrac{a^2}{x+a} = a - \dfrac{a^2}{x+a}$$ $\endgroup$
    – user17762
    Jun 18, 2012 at 2:15
  • $\begingroup$ Interesting, I have to admit I am not clear as to why the function is manipulated in this way. $\endgroup$
    – Kurt
    Jun 18, 2012 at 4:11
  • $\begingroup$ @Kurt You have a fraction with a polynomial on the top and on the bottom. You can generally get it into a form where there is a polynomial (perhaps zero or a constant) plus a fraction where the degree of the polynomial on top (the numerator) is less than the degree of the polynomial on the bottom (denominator). This is achieved by using the division algorithm for polynomials. $\endgroup$
    – Mark Bennet
    Jun 18, 2012 at 7:34
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$$ (a^{-1}+x^{-1})^{-1}=\frac{1}{a^{-1}+x^{-1}}=\frac{1}{\frac{1}{a}+\frac{1}{x}} = \frac{1}{\frac{x}{ax}+\frac{a}{ax}}= \frac{1}{\frac{x+a}{ax}}= \frac{ax}{x+a} $$

The horizontal asymptotes:

$$\lim_{x \to \infty} \frac{ax}{x+a}=\lim_{x \to -\infty} \frac{ax}{x+a}=a$$

i.e. at $y=a$

The vertical asymptotes:

$$\lim_{x \to -a^+} \frac{ax}{x+a} = \infty$$ $$\lim_{x \to -a^-} \frac{ax}{x+a} = -\infty$$

i.e. at $x=-a$

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    $\begingroup$ I found this explanation very clear and straight forward. To calculate horizontal asymptote we determine the limit as x approaches infinity, and to determine if there is vertical asymptote we calculate a value for $a$ where the function is undefined. $\endgroup$
    – Kurt
    Jun 18, 2012 at 4:18

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