Problem 5 - Precalculus

Question

Discuss/Explain why \(\lim _{x \rightarrow 3} g(x)=5\) for \(g(x)=\) \(\frac{x^{2}-x-6}{x-3},\) but \(g(3) \neq 5 .\) Can we redefine \(g(x)\) to create a piecewise-defined function where \(\lim _{x \rightarrow 3} g(x)=g(3) ?\)

Step-by-Step Solution

Answer

Redefine \(g(x)\) piecewise: \(g(x) = x+2\) for \(x \neq 3\) and \(g(3) = 5\); this aligns \(\lim _{x \rightarrow 3} g(x)\) with \(g(3)\).

Step 1: Understanding the Limit

The limit \(\lim _{x \rightarrow 3} g(x) = 5\) indicates that as \(x\) approaches 3, the function \(g(x)\) approaches 5. This must occur even if \(g(3)\) itself is not 5, often because at \(x=3\) the function \(g(x)\) is undefined or takes a different value.

Step 2: Simplifying the Function

First, express \(g(x) = \frac{x^{2}-x-6}{x-3}\). The polynomial in the numerator can be factored as \((x-3)(x+2)\). Therefore, \(g(x) = \frac{(x-3)(x+2)}{x-3}\). For \(x eq 3\), this simplifies to \(x+2\). However, \(x=3\) causes a division by zero, so \(g(3)\) is undefined at this point.

Step 3: Evaluating the Limit

With \(g(x)\) simplified to \(x+2\) for \(x eq 3\), calculate the limit \(\lim _{x \rightarrow 3} g(x) = \lim _{x \rightarrow 3} (x+2) = 3+2 = 5\). This confirms that \(\lim _{x \rightarrow 3} g(x) = 5\), as \(g(x)\) approaches 5 from both sides of \(x=3\).

Step 4: Evaluating \(g(3)\)

Although the expression \(g(x) = \frac{(x-3)(x+2)}{x-3}\) simplifies to \(x+2\) for \(x eq 3\), the original definition with the denominator \(x-3\) causes \(g(3)\) to be undefined. Thus, \(g(3)\) is not 5, it does not exist.

Step 5: Redefining the Function

To make \(g(3) = 5\), redefine \(g(x)\) in a piecewise manner such that \(g(x) = x+2\) when \(x eq 3\) and \(g(3) = 5\). Hence, the function now correctly aligns with its limit: \[ g(x) = \begin{cases} x+2, & \text{if } x eq 3 \ 5, & \text{if } x = 3 \end{cases} \]

Step 6: Conclusion

By redefining \(g(x)\) using the piecewise function, \(g(3)\) is now aligned with the limit \(\lim _{x \rightarrow 3} g(x)\), which is 5. This resolves the discontinuity present in the original function.

Key Concepts

Piecewise-defined function

A piecewise-defined function is one that has different expressions or rules for different intervals on its domain. It allows us to define a function that behaves differently at certain points while maintaining a consistent expression elsewhere. This flexibility is crucial when addressing issues such as discontinuities. In this context, the original function was redefined to ensure that the limit as we approach a specific point matches the function's value at that point. To create a piecewise-defined function, you map out two or more scenarios on how the function should behave:

• One scenario for values other than the problematic point, using the expression that simplifies correctly.
• Another scenario directly assigning a value at the problematic point to match the limit.

This curative method effectively aligns the behavior of the function across its domain with its limits. Consequently, any discontinuities can be smoothed out by clearly specifying what happens at these critical points.

Discontinuity in functions

Discontinuities in functions occur when a function is not continuous at certain points. Specifically, a function is continuous at a point if the limit at that point is equal to the function's value there. If they are not equal, or the value does not exist, a discontinuity exists. There are various types of discontinuities:

• Removable discontinuities: Where a limit exists, but the function is undefined at a specific point.
• Jump discontinuities: Where different one-sided limits exist and are not equal.
• Infinite discontinuities: Where limits approach infinity at the point.

In the original exercise, we dealt with a removable discontinuity. The function was undefined at one particular point due to division by zero, yet the limit exists as we approach that point from either direction. By redefining the function using a piecewise form, we resolved this discontinuity, thus making the function continuous by specifying its value explicitly across its entire domain.

Factorization of polynomials

Factorization of polynomials involves breaking down a complex polynomial into simpler polynomials that, when multiplied together, return the original polynomial. This process is particularly useful for simplifying functions and resolving issues like discontinuities. The given polynomial in the exercise, \(x^2 - x - 6\), can be expressed as a product of two simpler binomials: \((x-3)(x+2)\).Steps to factor a polynomial:

• Identify common factors, if any.
• Look for patterns, like difference of squares or trinomial squares.
• Apply the quadratic formula, if necessary, to find roots.
• Rewrite the polynomial as a product of its roots.

Through factorization, we can simplify \(g(x)\) to \(x + 2\) when \(x eq 3\), eliminating the term that causes division by zero. This simplifies the process of finding limits and helps redefine the function in a piecewise manner. Consequently, factorization not only aids in resolving algebraic expressions but also in handling functional discontinuities that might arise due to previous complexities in the expression.