What is the horizontal asymptote of $f(x)=\frac{-2 x}{x+1}$ ?
To determine the horizontal asymptote of the rational function \(f(x) = \frac{-2x}{x+1}\), we examine its behavior as \(x\) approaches positive or negative infinity. A horizontal asymptote exists if the limit of the function as \(x \to \infty\) or \(x \to -\infty\) results in a finite constant value.
We calculate the limit of \(f(x)\) as \(x\) approaches infinity:
\[\lim_{x \to \infty} \frac{-2x}{x+1}\]
To evaluate this limit, we divide both the numerator and the denominator by the highest power of \(x\) appearing in the denominator, which is \(x^1\):
\[\lim_{x \to \infty} \frac{\frac{-2x}{x}}{\frac{x+1}{x}} = \lim_{x \to \infty} \frac{-2}{1 + \frac{1}{x}}\]
As \(x \to \infty\), the term \(\frac{1}{x}\) approaches 0. Substituting this into the expression, we get:
\[\frac{-2}{1 + 0} = -2\]
Since the limit of the function as \(x\) approaches infinity is \(-2\), the horizontal line that the function approaches is \(y = -2\). Therefore, the horizontal asymptote of the given function is \(y = -2\).
In electrical engineering, the behavior of certain circuits can be modeled by rational functions where the voltage or current values approach a steady-state value over time. Identifying the horizontal asymptote allows engineers to determine the long-term, stable behavior of the system. For instance, in a system where the output voltage is modeled by \(V(t) = \frac{-2t}{t+1}\), the horizontal asymptote \(V = -2\) represents the steady-state voltage the circuit will reach as \(t\) becomes very large.