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? 

\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. $x=-a$.

