Problem 78 - Calculus: Early Transcendentals

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 \(\lim _{x \rightarrow a^{+}} f(x)\). $$f(x)=\frac{\left|1-x^{2}\right|}{x(x+1)}$$

Step-by-Step Solution

Answer

b) What are the vertical asymptotes of the function, and how does the function behave around them? Answer: a) The horizontal asymptote of the function is at \(y = 0\). b) The vertical asymptotes are at \(x = -1\) and \(x = 0\). Around the vertical asymptote \(x = -1\), the function is undefined. For the vertical asymptote \(x = 0\), the function goes to \(-\infty\) as \(x\) approaches 0 from the left and \(\infty\) as \(x\) approaches 0 from the right.

Step 1: Analyze \(\lim_{x \rightarrow \infty} f(x)\) and \(\lim_{x \rightarrow -\infty} f(x)\)

To find the horizontal asymptotes, we need to consider the limits as \(x\) approaches \(\infty\) and \(-\infty\). For \(\lim_{x \rightarrow \infty} f(x)\), we have: $$f(x) = \frac{|1-x^2|}{x(x+1)} = \frac{-x^2+1}{x(x+1)}$$ As \(x\) approaches infinity, the term \(-x^2+1\) will become increasingly large and negative, whereas both \(x\) and \((x+1)\) will become increasingly large and positive. Hence, the limit becomes: $$\lim_{x \rightarrow \infty} f(x) = \frac{-\infty}{+\infty} = 0$$ For \(\lim_{x \rightarrow -\infty} f(x)\), we have: $$f(x) = \frac{|1-x^2|}{x(x+1)} = \frac{x^2-1}{x(x+1)}$$ As \(x\) approaches negative infinity, the term \(x^2-1\) will become increasingly large and positive, whereas both \(x\) and \((x+1)\) will become increasingly large and negative. Hence, the limit becomes: $$\lim_{x \rightarrow -\infty} f(x) = \frac{+\infty}{+\infty} = 0$$ Since both limits approach the same value of 0, there is a horizontal asymptote at \(y=0\).

Step 2: Find the vertical asymptotes

Vertical asymptotes happen when the denominator of a function is equal to 0. In this case, the denominator is \(x(x+1)\). To find the vertical asymptotes, we must find the values of \(x\) for which the denominator is equal to 0. $$x(x+1) = 0$$ This equation has two solutions, \(x=-1\) and \(x=0\). These are the vertical asymptotes of the function.

Step 3: Analyze the limits for each vertical asymptote

Now we need to analyze the limits as \(x\) approaches each vertical asymptote from the left (\(x \rightarrow a^{-}\)) and the right (\(x \rightarrow a^{+}\)). For \(x=-1\), we have the following limits: $$\lim_{x \rightarrow -1^{-}} f(x) = \frac{|1-(-1)^2|}{(-1)(-1+1)} = \frac{0}{0} \Rightarrow \text{Undefined}$$ $$\lim_{x \rightarrow -1^{+}} f(x) = \frac{|1-(-1)^2|}{(-1)(-1+1)} = \frac{0}{0} \Rightarrow \text{Undefined}$$ For \(x=0\), we have the following limits: $$\lim_{x \rightarrow 0^{-}} f(x) = \frac{|1-(0)^2|}{(0)(0+1)} = \frac{1}{0} \Rightarrow -\infty$$ $$\lim_{x \rightarrow 0^{+}} f(x) = \frac{|1-(0)^2|}{(0)(0+1)} = \frac{1}{0} \Rightarrow \infty$$ So, we have vertical asymptotes at \(x=-1\) and \(x=0\), but the behavior around these asymptotes is different. In case of \(x=-1\), the function is undefined, and for \(x=0\), the function goes to \(-\infty\) as \(x\) approaches 0 from the left and \(\infty\) as \(x\) approaches 0 from the right.

Key Concepts

horizontal asymptotes

Horizontal asymptotes are lines that a graph approaches as the input, or x-value, moves towards positive or negative infinity. They help in understanding the end behavior of a function. For rational functions like \( f(x) = \frac{\left|1-x^{2}\right|}{x(x+1)} \), horizontal asymptotes can often be found by calculating the limit of \( f(x) \) as \( x \to \infty \) and \( x \to -\infty \).
To do this, we evaluate the highest power terms in the numerator and denominator, which typically dominate the function's behavior at extreme values of \( x \). In this example, you begin by rewriting \( f(x) \) appropriately for each limit scenario:

• As \( x \to \infty \), the function roughly simplifies to \( \frac{-x^2}{x^2} \), leading to \( \lim_{x \to \infty} f(x) = 0 \).
• As \( x \to -\infty \), the expression becomes \( \frac{x^2}{x^2} \), thus \( \lim_{x \to -\infty} f(x) = 0 \).

Both limits show that \( y = 0 \) is a horizontal asymptote. This means that regardless of how large or small \( x \) becomes, the function levels out towards \( y = 0 \). This asymptote provides insight into the overall shape and orientation of the graph.

vertical asymptotes

Vertical asymptotes occur at values of \( x \) where the function is undefined and "blows up," meaning it tends to infinity or negative infinity. For rational functions like \( f(x) = \frac{|1-x^2|}{x(x+1)} \), vertical asymptotes are often found at x-values that make the denominator zero, because division by zero is undefined.
To identify these asymptotes, solve \( x(x+1) = 0 \):

• This gives roots \( x = 0 \) and \( x = -1 \), indicating potential vertical asymptotes.

However, discovering the asymptotes is only half the battle; analyzing the behavior of the function around them is equally crucial:

• For \( x = 0 \), we see differing behavior from the left and right sides, with the limits leading to \( -\infty \) and \( \infty \) respectively, confirming a vertical asymptote.
• At \( x = -1 \), the limits approach an undefined value \( \frac{0}{0} \), indicating indeterminate form, but it signifies discontinuity at that point.

These speculations show that the graph goes towards extreme values near these lines, disrupting the continuity of the graph at \( x = -1 \) and \( x = 0 \). Vertical asymptotes are therefore a pivotal component of a function's graphical representation and help diagnose "gaps" or "breaks" in the curve.

limits in calculus

Limits are foundational in calculus, serving as a core concept to understand the behavior of functions as a variable approaches a certain value. When it comes to asymptotes, limits become crucial to predict the tendencies of a function at the edges of its domain.
The limit \( \lim_{x \to c} f(x) \) describes what happens to \( f(x) \) as \( x \) comes ever closer to \( c \), without necessarily reaching \( c \). For horizontal asymptotes, the focus is on \( \lim_{x \to \infty} f(x) \) and \( \lim_{x \to -\infty} f(x) \), elucidating how the function behaves as \( x \) stretches indefinitely in either direction.
In the context of vertical asymptotes, the limits \( \lim_{x \to a^-} f(x) \) and \( \lim_{x \to a^+} f(x) \) are often calculated to examine the function's approach from the left and right sides of the asymptote. These one-sided limits can reveal whether \( f(x) \) shoots up to \( +\infty \) or \( -\infty \) or if there's any indeterminate behavior to note.

• For example, \( \lim_{x \to 0^-} f(x) = -\infty \) and \( \lim_{x \to 0^+} f(x) = \infty \) show the contrasting behavior around \( x = 0 \), confirming a vertical asymptote.
• Such limits let us peek into the function's approach, offering a detailed picture of its trajectory.

In essence, calculating limits helps forecast how a function bends, breaks, or smooths out in its graph. Limits are indeed a significant stepping-stone in calculus, preparing students for more advanced topics like derivatives and integrals.