Question

a. Evaluate \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x),\) and then identify any horizontal asymptotes. b. Find the vertical asymptotes. For each vertical asymptote \(x=a\), evaluate \(\lim _{x \rightarrow a^{-}} f(x)\) and \(\lim _{x \rightarrow a^{+}} f(x)\). $$f(x)=\frac{x-1}{x^{2 / 3}-1}$$

Short answer

Question: Determine the horizontal and vertical asymptotes of the function \(f(x)=\frac{x-1}{x^{2/3}-1}\). Answer: The function has a horizontal asymptote at y = 0. There is no vertical asymptote.

Step-by-step solution

Part (a): Evaluate Limits & Identify Horizontal Asymptotes

To find the horizontal asymptotes, we will study the limits of the function as x approaches positive and negative infinity: 1. \(\lim_{x\rightarrow\infty}f(x)\) 2. \(\lim_{x\rightarrow-\infty}f(x)\) Consider the given function: $$f(x)=\frac{x-1}{x^{2/3}-1}$$ We first find the limit as x approaches positive infinity: To evaluate the limit, we can divide both the numerator and the denominator by the highest power of x, in this case, x. Then, we simplify and evaluate the limit: $$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}\frac{\frac{x-1}{x}}{\frac{x^{2/3}-1}{x}}=\lim_{x\rightarrow\infty}\frac{1-\frac{1}{x}}{x^{1/3}-\frac{1}{x}}$$ As x approaches infinity, \(\frac{1}{x}\) approaches 0, and hence the limit becomes $$\lim_{x\rightarrow\infty}\frac{1-\frac{1}{x}}{x^{1/3}-\frac{1}{x}}=\frac{1}{0}= \infty$$ Now we find the limit as x approaches negative infinity: $$\lim_{x\rightarrow-\infty}f(x)=\lim_{x\rightarrow-\infty}\frac{x-1}{x^{2/3}-1}$$ The same method as before: $$\lim_{x\rightarrow-\infty}f(x)=\lim_{x\rightarrow-\infty}\frac{\frac{x-1}{x}}{\frac{x^{2/3}-1}{x}}=\lim_{x\rightarrow-\infty}\frac{1-\frac{1}{x}}{x^{-1/3}-\frac{1}{x}}$$ As x approaches negative infinity, \(\frac{1}{x}\) approaches 0, and hence the limit becomes $$\lim_{x\rightarrow-\infty}\frac{1-\frac{1}{x}}{x^{-1/3}-\frac{1}{x}}=\frac{1}{\infty}=0$$ So, there is a horizontal asymptote at y = 0.

Part (b): Find Vertical Asymptotes and Limits

To find vertical asymptotes, we must identify the undefined values of x in the function. i.e., when the denominator is equal to 0. $$x^{2/3}-1=0$$ We can rewrite this equation as: $$x^{2/3}=1$$ Taking the cube root of both sides, we get: $$x=1$$ So, x = 1 is the vertical asymptote. Now, we will evaluate the limits of the function as it approaches 1 from the left and right: 1. \(\lim_{x\rightarrow1^{-}} f(x)\) 2. \(\lim_{x\rightarrow1^{+}} f(x)\) To approach 1 from the left and right, we can use direct substitution. $$\lim_{x\rightarrow1^{-}} f(x)=\lim_{x\rightarrow1^{+}} f(x)=f(1)$$ Substituting x = 1 into the function, we get: $$f(1)=\frac{1-1}{1^{2/3} - 1}=0$$ Since the limit exists and it's equal to 0, there is no vertical asymptote at x = 1. To summarize: a. There is a horizontal asymptote at y = 0. No horizontal asymptote exists when x approaches infinity. b. No vertical asymptote exists at x = 1.

Key concepts

Limits

Limits are fundamental in understanding the behavior of functions as they approach specific values or infinity. They help us analyze the behavior of a function, especially when it can't be easily evaluated through direct substitution. In today's exercise, we are focusing on limits as they approach infinity and how they help identify horizontal asymptotes.

The given function is \( f(x) = \frac{x-1}{x^{2/3}-1} \), and the task is to find the limits as \( x \) approaches both positive and negative infinity. By dividing both the numerator and the denominator by the highest degree of \( x \), we can simplify it for limit evaluation:

• As \( x \rightarrow \infty \), it simplifies to \( \lim_{x\rightarrow\infty}\frac{1-\frac{1}{x}}{x^{1/3}-\frac{1}{x}} \).
• As \( x \rightarrow -\infty \), it similarly simplifies to \( \lim_{x\rightarrow-\infty}\frac{1-\frac{1}{x}}{x^{-1/3}-\frac{1}{x}} \).

As \( \frac{1}{x} \) approaches 0 when \( x \) is extremely large or small, it helps calculate the function's end behavior.

Horizontal Asymptotes

Horizontal asymptotes reveal the ultimate behavior of a function as it stretches towards infinity. They indicate where the function 'settles' as \( x \) approaches infinity or negative infinity.

For the function \( f(x) = \frac{x-1}{x^{2/3}-1} \), we use limits at infinity to detect horizontal asymptotes. Analyzing the limits we calculated:

• \( \lim_{x\rightarrow\infty} f(x) = \infty \) means there is no horizontal asymptote in the positive direction.
• \( \lim_{x\rightarrow-\infty} f(x) = 0 \) indicates a horizontal asymptote at \( y = 0 \) in the negative direction.

Thus, the horizontal asymptote aligns with the line \( y = 0 \), showing the function's tendencies towards negative infinity.

Vertical Asymptotes

Vertical asymptotes point to values where a function becomes undefined—a place where the denominator is zero, making the function's value shoot up to infinity or down to negative infinity.

To locate these asymptotes in \( f(x) = \frac{x-1}{x^{2/3}-1} \), we solve when the denominator equals zero: \( x^{2/3} - 1 = 0 \), leading to \( x = 1 \).

Next, evaluate the function close to this potential asymptote by considering:

• \( \lim_{x\rightarrow1^{-}} f(x) \)
• \( \lim_{x\rightarrow1^{+}} f(x) \)

By using substitution, the function \( \frac{0}{0} \) surprisingly simplifies to 0 here, declaring no vertical asymptote at \( x = 1 \). This is an interesting twist, reminding us that not all zeros in the denominator lead to asymptotes, particularly when they cancel with the numerator.