Problem 87 - Calculus

Question

In Problems \(87-88,\) give an example of: A rational function that has a limit at \(x=1\) but is not continuous at \(x=1\)

Step-by-Step Solution

Answer

The function \(f(x) = \frac{x^2-1}{x-1}\) has a limit as \(x\to1\) but is not continuous at \(x=1\).

Step 1: Understand the Problem

We need to give an example of a rational function, \(f(x) = \frac{P(x)}{Q(x)}\), that has a limit as \(x\) approaches 1, but is not continuous at \(x=1\). This means \(\lim_{x \to 1} f(x)\) exists but \(f(x)\) is undefined at \(x=1\).

Step 2: Identify the Conditions for Continuity

For a function to be continuous at \(x=1\), three conditions must be met:1. \(f(1)\) is defined.2. The limit \(\lim_{x \to 1} f(x)\) exists.3. \(\lim_{x \to 1} f(x) = f(1)\).To fail continuity, one of these conditions must fail. We will ensure that \(f(1)\) is undefined by making \(Q(x) = 0\) when \(x=1\).

Step 3: Choose a Rational Function Form

Select a rational function \(f(x) = \frac{P(x)}{Q(x)}\) such that \(Q(x)\) has a factor \((x-1)\) in it. This will ensure that \(f(x)\) is undefined at \(x=1\). A simple choice would be \(Q(x) = x-1\).

Step 4: Ensure the Limit Exists

To make sure the limit exists as \(x\) approaches 1, ensure that the function simplifies as \(x\) nears 1. A reasonable choice is to have \(P(x) = x^2-1\). Factor \(P(x)\) as \((x-1)(x+1)\).

Step 5: Define the Function

Let \(f(x) = \frac{x^2-1}{x-1}\). This reduces to \(g(x) = x+1\) when \(x eq 1\).

Step 6: Check the Limit and Discontinuity

Calculate \(\lim_{x \to 1} \frac{x^2-1}{x-1}\) by substituting \(x \to 1\) in \(g(x) = x+1\), giving us \(2\). However, \(f(x)\) is not continuous at \(x=1\) as it is undefined due to the denominator being zero at \(x=1\).

Step 7: Provide Example Function

The function \(f(x) = \frac{x^2-1}{x-1}\) meets the conditions: - It simplifies to \(g(x) = x+1\), making \(\lim_{x \to 1} f(x) = 2\).- It is undefined at \(x=1\), so it is not continuous there.

Key Concepts

Understanding Continuity in Rational Functions

In mathematics, a function is said to be continuous at a specific point when there is no interruption, jump, or gap in its graph at that point. Continuity demands that three conditions be fulfilled for a rational function, such as the one given by \( f(x) = \frac{P(x)}{Q(x)} \), at a point like \( x=1 \):

• \( f(1) \) is defined: The function has a valid output at \( x=1 \).
• \( \lim_{x \to 1} f(x) \) exists: The input values approaching \( x=1 \) from either direction lead to the same limit.
• \( \lim_{x \to 1} f(x) = f(1) \): The limit matches the function value at \( x=1 \).

When a function is continuous at every point in its domain, it is said to be continuous everywhere in that domain.
The concept of continuity is crucial as it ensures the function behaves predictably around that point, without any unexpected jumps or breaks.

Exploring Limits in Rational Functions

Limits help us to understand the behavior of a function as the input approaches a certain point, even if the function is not defined exactly at that point. For rational functions, which are fractions of polynomials, we look at what happens as \( x \) approaches a specific number, such as 1 in the example of \( f(x) = \frac{x^2-1}{x-1} \).
To find the limit \( \lim_{x \to 1} \frac{x^2-1}{x-1} \), we simplify the expression first. The polynomial in the numerator \( x^2 - 1 \) can be factored as \( (x-1)(x+1) \). This allows the \( (x-1) \) factors in the numerator and denominator to cancel out (for \( x eq 1 \)), simplifying the function to \( g(x) = x+1 \).

• By evaluating \( g(x) \), we find that as \( x \to 1 \), \( g(x) \to 2 \).
• This shows the limit of the original rational function is 2 as \( x \) approaches 1, even though the function isn't defined at \( x=1 \).

Understanding limits helps in analyzing and predicting the value trends of functions near points of interest.

Identifying Discontinuity in Rational Functions

Discontinuity occurs when a function fails to meet the conditions of continuity at certain points in its domain. With rational functions, these discontinuities often arise due to the denominator equalling zero, leading the function to be undefined there.
In the step-by-step solution, the function \( f(x) = \frac{x^2-1}{x-1} \) furnishes an example of this. While \( \lim_{x \to 1} f(x) = 2 \) suggests continuity, \( f(x) \) itself is not defined at \( x=1 \). Here's why:

• The denominator \( x-1 \) results in zero at \( x=1 \), causing the function can't produce a value.
• The presence of an undefined value contradicts the first condition of continuity, producing a removable discontinuity.

This type of discontinuity, also known as a hole, is visible in the graph of \( f(x) \), where there is a gap at \( x=1 \). Learning to detect and describe these discontinuities is vital for understanding the behavior and limits of rational functions.