Problem 70 - 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 \(\lim _{x \rightarrow a^{+}} f(x)\). $$f(x)=\frac{x^{2}-4 x+3}{x-1}$$

Step-by-Step Solution

Answer

Question: Determine the horizontal and vertical asymptotes of the given rational function, if any: $$f(x)=\frac{x^{2}-4 x+3}{x-1}$$ Answer: There are no horizontal and vertical asymptotes for the given rational function because it has a removable discontinuity at x=1.

Step 1: Find the limits as x approaches infinity and negative infinity

First, we need to analyze the following two limits to determine if any horizontal asymptotes exist: $$\lim _{x \rightarrow \infty} f(x)\text{ and }\lim _{x \rightarrow-\infty} f(x)$$ For the given function: $$f(x)=\frac{x^{2}-4 x+3}{x-1}$$ As x becomes very large (either positive or negative), \(x^2\) will dominate the numerator, and x will dominate the denominator. Thus, we can simplify the expression and analyze the resulting limit: $$\lim_{x \rightarrow \pm\infty}\frac{x^{2}-4 x+3}{x-1} \sim \lim_{x \rightarrow \pm\infty}\frac{x^2}{x}$$ Now, divide both terms by x, and we get: $$\lim_{x \rightarrow \pm\infty}\frac{x^{2}}{x}= \lim_{x \rightarrow \pm\infty}x$$ Since x goes to infinity or negative infinity, there are no horizontal asymptotes.

Step 2: Find the vertical asymptotes

To find the vertical asymptotes, we need to determine the values of x for which the denominator becomes zero. So, let's set the denominator equal to zero: $$x - 1 = 0$$ Solving for x, we get: $$x = 1$$ So there is a vertical asymptote at \(x = 1\).

Step 3: Analyze the limits as x approaches the vertical asymptote

Now, we need to analyze the limits as x approaches the vertical asymptote from both the left side (\(x \rightarrow a^-\)) and the right side (\(x \rightarrow a^+\)). In this case, \(a = 1\). Let's find the two limits: $$\lim _{x \rightarrow 1^{-}} f(x)\text{ and }\lim _{x \rightarrow 1^{+}} f(x)$$ Using the given function: $$f(x)=\frac{x^{2}-4 x+3}{x-1}$$ Factor the numerator: $$f(x)=\frac{(x-3)(x-1)}{x-1}$$ We can see that there is a removable discontinuity at x=1. So, we can factor out \((x-1)\) in both the numerator and the denominator and simplify the function to: $$f(x)= x-3$$ Then, analyzing the limits as x approaches 1 from both directions, $$\lim _{x \rightarrow 1^{-}} (x-3) = \lim _{x \rightarrow 1^{+}} (x-3) = -2$$ Since the function has a removable discontinuity at x=1, but no infinite limits as x approaches 1 from either side, there is no vertical asymptote.

Step 4: Final answer

We found that there are no horizontal and vertical asymptotes for the given rational function because it has a removable discontinuity at x=1.

Key Concepts

Horizontal Asymptotes

Horizontal asymptotes are lines that the graph of a function approaches as the input variable becomes very large (positively or negatively). To identify horizontal asymptotes, we need to evaluate the behavior of the function as the input grows towards infinity or negative infinity. In our example, the function is given by \(f(x)=\frac{x^{2}-4x+3}{x-1}\).

For large \{x}\ values, the term\{x^2}\ becomes the most significant in the numerator, while \{x}\ itself dominates in the denominator. By simplifying, you'll have \(f(x) \approx \frac{x^{2}}{x} = x\).

Since \{x}\ marches toward infinity (or negative infinity), the function \(f(x)\) does not level off to approach a constant value. Therefore, the function exhibits no horizontal asymptotes.

Vertical Asymptotes

Vertical asymptotes are vertical lines that the graph of a function approaches but never crosses. They occur at points where a function tends toward infinity. This usually happens when the function's denominator equals zero, causing the function to blow up.

For the function \(f(x)=\frac{x^{2}-4x+3}{x-1}\), to find possible vertical asymptotes, we set the denominator to zero: \{x-1=0}\, giving us \{x=1}\.

Initially, this seems like an asymptote at \{x=1}\, but we need more analysis. When we factor the function's numerator, we find \(f(x)=\frac{(x-3)(x-1)}{x-1}\). After canceling out \(x-1\) in the numerator and denominator, we are left with \(f(x)=x-3\), which simplifies to a linear function without a vertical asymptote because the function does not approach infinity.

Limits at Infinity

The concept of limits at infinity helps us understand the behavior of a function as \{x}\ approaches infinity or negative infinity. For rational functions like \(f(x) = \frac{x^{2}-4x+3}{x-1}\), the limits at infinity help determine the existence of horizontal asymptotes.

We analyze \{\lim_{x \to \infty} f(x)}\ and \{\lim_{x \to -\infty} f(x)}\ by focusing on the highest powers in the numerator and denominator. As we simplify \(f(x) \approx \frac{x^{2}}{x} = x\), the absence of a constant limit means that \(f(x)\) doesn't approach a horizontal line, indicating no horizontal asymptotes are present.

Therefore, even as \{x}\ becomes very large in magnitude, the function continues to grow or decrease without leveling off.

Removable Discontinuity

A removable discontinuity occurs when a function is undefined at a certain point, but approaches the same finite limit from either side of that point. This is usually due to common factors in the numerator and denominator that can be canceled.

For the function \(f(x)=\frac{x^{2}-4x+3}{x-1}\), factoring the numerator gives \(f(x)=\frac{(x-3)(x-1)}{x-1}\). Canceling \{x-1}\ from the numerator and denominator shows the function simplifies to \(f(x)=x-3\), except at \{x=1}\.

So, \(x=1\) is a removable discontinuity—not a vertical asymptote—since the function simplifies and approaches \{-2}\ from both sides. This means the function is undefined at exactly \{x=1}\, but behaves regularly just to the left and right of this value.