Problem 57 - Calculus Early Transcendentals

Question

a. Analyze \(\lim f(x)\) 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 \(\lim _{x \rightarrow a^{+}} f(x)\) $$f(x)=\frac{x^{2}-9}{x(x-3)}$$

Step-by-Step Solution

Answer

Answer: The horizontal asymptote of the function is \(y=1\), and the vertical asymptote is \(x=0\). As the function approaches the vertical asymptote from the left, the limit is negative infinity, and as it approaches from the right, the limit is positive infinity.

Step 1: Factorize the expression

We can write the given function as: $$f(x)=\frac{(x+3)(x-3)}{x(x-3)}$$

Step 2: Simplify the function

Cancel the common factor of \((x-3)\) in both numerator and denominator: $$f(x)=\frac{x+3}{x}$$

Step 3: Find the limit as x approaches infinity

By dividing both numerator and denominator by x, we have: $$f(x)=\frac{\frac{x+3}{x}}{\frac{x}{x}}=\frac{1+\frac{3}{x}}{1}$$ Now, take the limit as x approaches infinity: $$\lim_{x \rightarrow+\infty}f(x)=\lim_{x \rightarrow+\infty}\frac{1+\frac{3}{x}}{1}$$ As \(x\) approaches infinity, \(\frac{3}{x}\) approaches 0, so the limit is 1: $$\lim_{x \rightarrow+\infty}f(x)=1$$

Step 4: Determine the horizontal asymptote

Since the limit as \(x\) approaches infinity is 1, the function will approach a horizontal line at \(y=1\). Thus, the horizontal asymptote is: $$y=1$$ #b. Finding the vertical asymptotes and analyzing the limits#

Step 5: Find the vertical asymptotes

Vertical asymptotes occur when the denominator of a rational function is equal to 0. In our simplified function, the denominator is \(x\). Therefore, the vertical asymptote is: $$x=0$$

Step 6: Analyze the limit as x approaches 0 from the left and right

First, let's find \(\lim_{x \rightarrow a^{-}} f(x)\) as \(x\) approaches 0: $$\lim_{x \rightarrow 0^{-}}f(x)=\lim_{x \rightarrow 0^{-}}\frac{x+3}{x}$$ As \(x\) approaches 0 from the left, the denominator approaches 0 from the negative side, so the limit is negative infinity: $$\lim_{x \rightarrow 0^{-}}f(x)=-\infty$$ Now, let's find \(\lim_{x \rightarrow a^{+}} f(x)\) as \(x\) approaches 0: $$\lim_{x \rightarrow 0^{+}}f(x)=\lim_{x \rightarrow 0^{+}}\frac{x+3}{x}$$ As \(x\) approaches 0 from the right, the denominator approaches 0 from the positive side, so the limit is positive infinity: $$\lim_{x \rightarrow 0^{+}}f(x)=+\infty$$ In conclusion: a. The limit as \(x\) approaches infinity is 1, and the horizontal asymptote is \(y=1\). b. The vertical asymptote is \(x=0\), with the function approaching negative infinity as \(x\) approaches 0 from the left, and positive infinity as \(x\) approaches 0 from the right.

Key Concepts

Horizontal Asymptotes

In mathematics, a horizontal asymptote is a horizontal line that a graph of a function approaches as the input or x-values approach infinity or negative infinity. Horizontal asymptotes describe the behavior of a function as it becomes very large or very small, giving us a boundary that the graph of the function approaches but never quite reaches.

To find horizontal asymptotes for the function \(f(x) = \frac{x^2 - 9}{x(x-3)}\), we look at the limits as \(x\) approaches infinity. By simplifying, the function becomes \(\frac{x+3}{x}\). Dividing each term by \(x\) leads us to the function \(f(x) = 1 + \frac{3}{x}\). As \(x\) approaches infinity, \(\frac{3}{x}\) becomes negligible, approaching zero. Hence, the horizontal asymptote is\(y=1\).

• Horizontal asymptotes help predict the long-term behavior of functions.
• They apply to rational functions where both the numerator and denominator are polynomials.

Vertical Asymptotes

Vertical asymptotes occur at the values of \(x\) that make the denominator of a rational function zero, indicating that the function is undefined at those points. These are the points where the function tends to positive or negative infinity, leading to a vertical line on the graph of the function.

For the function \(f(x) = \frac{x+3}{x}\), the denominator \(x\) becomes zero when \(x=0\). Thus, \(x=0\) is a vertical asymptote. As \(x\) approaches zero from the left (\(x\rightarrow0^{-}\)), the fraction \(\frac{x+3}{x}\) approaches negative infinity since dividing by a negative near-zero number results in a negatively large quotient. Conversely, as \(x\) approaches zero from the right (\(x\rightarrow0^{+}\)), the function approaches positive infinity.

• Vertical asymptotes show where a function increases or decreases without bound.
• They are particularly crucial in analyzing rational functions.

Rational Functions

Rational functions form an important category of functions in mathematics. They are expressed as the ratio of two polynomials, \(f(x) = \frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x)\) is not the zero polynomial. The domain of a rational function consists of all real numbers except those that make the denominator zero.

Rational functions frequently feature both horizontal and vertical asymptotes, which provide insight into their behavior. Horizontal asymptotes pertain to limits as \(x\) approaches plus or minus infinity, showing how the function behaves at extremes. Vertical asymptotes occur at zeroes of the denominator not canceled by the numerator, indicating points of infinite discontinuity.

• If the degree of the polynomial in the numerator is less than that of the denominator, \(y = 0\) is typically a horizontal asymptote.
• If the degrees are equal, the asymptote is \(y = \frac{a}{b}\), where \(a\) and \(b\) are the leading coefficients.