Question

Horizontal and vertical asymptotes. a. Analyze \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x),\) and then identify any horizontal asymptotes. b. Find the vertical asymptotes. For each vertical asymptote \(x=a\) analyze \(\lim _{x \rightarrow a^{-}} f(x)\) and \(\lim _{x \rightarrow a^{+}} f(x)\). $$f(x)=\frac{x^{2}-9}{x(x-3)}$$

Short answer

Answer: The horizontal asymptote is at \(y=1\). The vertical asymptotes are at \(x=0\) and \(x=3\). The limits as x approaches these asymptotes are: $$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=-3$$ $$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=\frac{1}{3}$$

Step-by-step solution

Analyze \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x)\)

To find horizontal asymptotes, let's compute the limits of \(f(x)\) as \(x\) approaches positive and negative infinity. $$\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty}\frac{x^{2}-9}{x(x-3)}$$ $$\lim _{x \rightarrow-\infty} f(x)=\lim _{x \rightarrow-\infty}\frac{x^{2}-9}{x(x-3)}$$

Calculate the limits

To find the limits, we can divide the numerator and denominator by the highest power of \(x\) in the denominator, which is \(x\), during this process we will obtain the simplified form of the given function: $$\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty}\frac{\frac{x^{2}}{x}-\frac{9}{x}}{\frac{x^{2}}{x}-\frac{3x}{x}}$$ $$\lim _{x \rightarrow-\infty} f(x)=\lim _{x \rightarrow-\infty}\frac{\frac{x^{2}}{x}-\frac{9}{x}}{\frac{x^{2}}{x}-\frac{3x}{x}}$$ Now, simplify the limits: $$\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty}\frac{x-\frac{9}{x}}{x-3}$$ $$\lim _{x \rightarrow-\infty} f(x)=\lim _{x \rightarrow-\infty}\frac{x-\frac{9}{x}}{x-3}$$ As \(x\) approaches infinity, the term \(\frac{9}{x}\) approaches zero. Therefore, the limits become: $$\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty}\frac{x}{x-3}=1$$ $$\lim _{x \rightarrow-\infty} f(x)=\lim _{x \rightarrow-\infty}\frac{x}{x-3}=1$$

Identify horizontal asymptotes

Since both limits approach the constant value of 1, we can conclude that there is a horizontal asymptote at \(y=1\). #b. Finding vertical asymptotes and analyzing their limits#

Find vertical asymptotes

To find the vertical asymptotes, we should look for the values of \(x\) that make the denominator of the function equal to zero: \(x(x-3)=0\) Solving this equation, we find two values for \(x\): \(x=0\) and \(x=3\) These values indicate vertical asymptotes at \(x=0\) and \(x=3\).

Analyze the limits of the function as x approaches the vertical asymptotes

Now we analyze the limits of the function as \(x\) approaches the vertical asymptotes from the left and right: $$\lim _{x \rightarrow 0^{-}} f(x)$$ $$\lim _{x \rightarrow 0^{+}} f(x)$$ $$\lim _{x \rightarrow 3^{-}} f(x)$$ $$\lim _{x \rightarrow 3^{+}} f(x)$$ Notice that as \(x\) approaches \(0\), the term \(\frac{x^{2}-9}{x-3}\) is approaching \(-3\) and thus: $$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=-3$$ As \(x\) approaches \(3\), the term \(\frac{x^{2}-9}{x}\) is approaching \(\frac{1}{3}\) and thus: $$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=\frac{1}{3}$$ In conclusion, the given function \(f(x)=\frac{x^{2}-9}{x(x-3)}\) has a horizontal asymptote at \(y=1\), vertical asymptotes at \(x=0\) and \(x=3\), and as \(x\) approaches these vertical asymptotes, the function behaves as follows: $$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=-3$$ $$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=\frac{1}{3}$$

Key concepts

Horizontal Asymptotes

When we talk about horizontal asymptotes, we are referring to the behavior of a function as the variable, usually denoted as \(x\), approaches either positive or negative infinity. In simpler terms, it tells us how the function behaves when \(x\) becomes extremely large or extremely large in the negative direction. Horizontal asymptotes are represented as horizontal lines, typically written as \(y = c\), where \(c\) is a constant value the function approaches but never actually reaches completely.

Let's consider the function \(f(x) = \frac{x^{2}-9}{x(x-3)}\). To find the horizontal asymptotes, we need to see what happens to \(f(x)\) as \(x\) tends to infinity or negative infinity. By calculating \(\lim_{x \to \infty} f(x)\) and \(\lim_{x \to -\infty} f(x)\), we can understand this behavior.

• We start by simplifying the expression by dividing by \(x\), the highest power in the denominator. It simplifies to \(\frac{x - \frac{9}{x}}{x - 3}\).
• As \(x\) grows very large, the term \(\frac{9}{x}\) becomes negligible, leaving us with \(\frac{x}{x-3}\).
• This fraction simplifies to \(1\), revealing that \(f(x)\) approaches \(1\).

As the limits at infinity and negative infinity yield \(1\), we can identify that \(y = 1\) is a horizontal asymptote of the function. In the graph, the function flattens around this line as \(x\) grows very large in either direction.

Vertical Asymptotes

Vertical asymptotes are lines where the function becomes unbounded or infinitely large, making the function undefined at those points. They're closely related to division by zero scenarios in the function's domain. For the function \(f(x) = \frac{x^{2}-9}{x(x-3)}\), a vertical asymptote will occur when the denominator equals zero because that makes the function undefined.

Performing a simple check on the denominator \(x(x-3) = 0\), we find:

• For \(x = 0\), the denominator becomes zero, hence indicating a vertical asymptote at this point.
• Similarly, for \(x = 3\), the denominator once again hits zero, signaling another vertical asymptote.

Therefore, we have vertical asymptotes at \(x = 0\) and \(x = 3\). Let's take a closer look at the behavior of the function at these asymptotes:
- As \(x\) approaches \(0\) from both sides, the function approaches negative infinity, confirming the vertical asymptote at \(x = 0\).
- Approaching \(x = 3\) from either direction results in the function values shooting towards positive or negative infinity, further solidifying \(x = 3\) as a vertical asymptote.

Limits

The concept of limits is foundational in calculus and helps us understand the behavior of functions at points where they might be undefined or as they approach infinity. Limits allow us to rigorously define the value that a function approaches as the input approaches a certain point, whether finite or infinite.

In the exercise, analyzing limits is crucial for identifying asymptotes of the function \(f(x) = \frac{x^{2}-9}{x(x-3)}\). For horizontal asymptotes, we compute limits as \(x\) approaches infinity and negative infinity, which show us the end-behavior pattern of the function. If these limits exist and are finite, they define the horizontal asymptote.

• For horizontal asymptotes, we computed \(\lim_{x \to \infty} f(x)\) and \(\lim_{x \to -\infty} f(x)\), both equaling \(1\), indicating \(y=1\) is the horizontal asymptote.
• Vertical asymptotes necessitate examining limits as the function approaches unspecified points, such as the roots of the denominator. Here, we focused on \(x = 0\) and \(x = 3\), confirming infinite behavior through evaluating \(\lim_{x \to 0^-} f(x)\), \(\lim_{x \to 0^+} f(x)\), \(\lim_{x \to 3^-} f(x)\), and \(\lim_{x \to 3^+} f(x)\).

By analyzing limits, we gain significant insights into a function's behavior, both at its boundaries and its undefined points, enabling a comprehensive understanding of where the function approaches or diverges.