Find the vertical and horizontal asymptotes of the function. 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? 
 A: 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 $y=a$.
A: \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. $x=-a$.

A: $$
(a^{-1}+x^{-1})^{-1}=\frac{1}{a^{-1}+x^{-1}}=\frac{1}{\frac{1}{a}+\frac{1}{x}} = 
\frac{1}{\frac{x}{ax}+\frac{a}{ax}}= 
\frac{1}{\frac{x+a}{ax}}=
\frac{ax}{x+a}
$$
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 $x=-a$
