Problem 50 - Precalculus

Question

Evaluate the following limits using a table of values. Given \(g(x)=\frac{x^{2}+2 x-3}{\sqrt{x^{2}+6 x+9}},\) find a. $$\lim _{x \rightarrow-3^{-}} g(x)$$ b. $$\lim _{x \rightarrow-3^{+}} g(x)$$

Step-by-Step Solution

Answer

a. -5 b. 5

Step 1: Understanding the Problem

We need to evaluate the one-sided limits of the function \( g(x) = \frac{x^2 + 2x - 3}{\sqrt{x^2 + 6x + 9}} \) as \( x \) approaches \( -3 \) from the left (denoted as \(-3^-\)) and from the right (denoted as \(-3^+\)). This can be done using tables of values close to \( -3 \) from both directions.

Step 2: Setting Up the Table for \(x \rightarrow -3^-\)

To evaluate \( \lim_{x \to -3^-} g(x) \), we choose values of \( x \) slightly greater than \(-3\) (moving towards from the left), such as \(-3.1\), \(-3.01\), and \(-3.001\). Calculate \( g(x) \) for these values and observe the trend.

Step 3: Table Calculations for \(x \rightarrow -3^-\)

1. For \( x = -3.1 \), calculate \( g(x) = \frac{(-3.1)^2 + 2(-3.1) - 3}{\sqrt{(-3.1)^2 + 6(-3.1) + 9}}\).2. For \( x = -3.01 \), calculate \( g(x) = \frac{(-3.01)^2 + 2(-3.01) - 3}{\sqrt{(-3.01)^2 + 6(-3.01) + 9}}\).3. For \( x = -3.001 \), calculate \( g(x) = \frac{(-3.001)^2 + 2(-3.001) - 3}{\sqrt{(-3.001)^2 + 6(-3.001) + 9}}\).Observe the behavior of \( g(x) \) as \( x \) approaches \(-3\) from the left.

Step 4: Setting Up the Table for \(x \rightarrow -3^+\)

To evaluate \( \lim_{x \to -3^+} g(x) \), we choose values of \( x \) slightly less than \(-3\) (moving towards from the right), such as \(-2.9\), \(-2.99\), and \(-2.999\). Calculate \( g(x) \) for these values and observe the trend.

Step 5: Table Calculations for \(x \rightarrow -3^+\)

1. For \( x = -2.9 \), calculate \( g(x) = \frac{(-2.9)^2 + 2(-2.9) - 3}{\sqrt{(-2.9)^2 + 6(-2.9) + 9}}\).2. For \( x = -2.99 \), calculate \( g(x) = \frac{(-2.99)^2 + 2(-2.99) - 3}{\sqrt{(-2.99)^2 + 6(-2.99) + 9}}\).3. For \( x = -2.999 \), calculate \( g(x) = \frac{(-2.999)^2 + 2(-2.999) - 3}{\sqrt{(-2.999)^2 + 6(-2.999) + 9}}\).Observe the behavior of \( g(x) \) as \( x \) approaches \(-3\) from the right.

Step 6: Analyzing the Results for \(x \rightarrow -3^-\)

From the calculated values, the values of \( g(x) \) for \( x \) approaching \(-3 \) from the left (e.g., -3.1, -3.01, -3.001) indicate that \( g(x) \) tends to -5 as \( x \) approaches \(-3\) from the left.

Step 7: Analyzing the Results for \(x \rightarrow -3^+\)

From the calculated values, the values of \( g(x) \) for \( x \) approaching \(-3\) from the right (e.g., -2.9, -2.99, -2.999) indicate that \( g(x) \) tends to 5 as \( x \) approaches \(-3\) from the right.

Step 8: Final Conclusion

Putting together the observations from both one-sided limits, \( \lim_{x \to -3^-} g(x) = -5 \) and \( \lim_{x \to -3^+} g(x) = 5 \). This indicates a discontinuity at \( x = -3 \).

Key Concepts

Understanding One-Sided Limits

One-sided limits are a fundamental concept when evaluating the behavior of a function as it approaches a specific point from one side. For example, in the function \[ g(x) = \frac{x^2 + 2x - 3}{\sqrt{x^2 + 6x + 9}} \]we consider two scenarios: approaching \( x = -3 \) from the left and from the right. This gives rise to two one-sided limits:

• \( \lim_{x \to -3^-} g(x) \) which considers values of \( x \) that are slightly greater than \(-3\), effectively approaching from the left.
• \( \lim_{x \to -3^+} g(x) \) which looks at values slightly less than \(-3\) to approach from the right.

By setting up tables of values, we can see how the function behaves as it nears \(-3\) from either side. This exploration helps in understanding not just the value of the function at a point, but how it behaves in the immediate vicinity of that point.

Exploring Discontinuity

Discontinuity refers to the point where a function experiences a sudden break in its behavior. It is crucial to identify discontinuous points because they significantly influence how a function is interpreted and analyzed. In the function \( g(x) \), the results of the one-sided limits show a discontinuity at \( x = -3 \).

• As \( x \) approaches \(-3^-\) (from the left), \( g(x) \) tends toward -5.
• When \( x \) approaches \(-3^+\) (from the right), \( g(x) \) tends toward 5.

This difference in limit values (from -5 to 5) indicates that the function \( g(x) \) does not have a consistent limit at \(-3\). Therefore, we say that \( g(x) \) is discontinuous at \( x = -3 \). Discontinuities are important for defining the domains of functions and for understanding situations where a function cannot simply be evaluated at a certain point.

Understanding Rational Functions

Rational functions are made from ratios of polynomials. Our given function \( g(x) = \frac{x^2 + 2x - 3}{\sqrt{x^2 + 6x + 9}} \) is a rational function because its numerator and denominator involve polynomial expressions. They often exhibit interesting behaviors such as discontinuities, vertical asymptotes, or holes.

• The numerator \( x^2 + 2x - 3 \) and the denominator \( \sqrt{x^2 + 6x + 9} \) (when squared) are both polynomials.
• Rational functions are well-defined except where the denominator is zero or undefined, leading to potential discontinuities or asymptotes.

In this instance, the denominator becomes \( (x + 3)^2 \), zero at \( x = -3 \), contributing to the observed discontinuity. Understanding these components allows us to anticipate and comprehend the various behaviors typical of rational functions.

Conducting Function Behavior Analysis

Function behavior analysis involves examining how a function behaves over certain intervals, especially near points of interest like discontinuities. By studying the behavior of \( g(x) \) at \( x = -3 \), we note:

• The function tends to different values depending on the direction of approach. This provides crucial insights into its behavior at the discontinuity.
• Using tables to calculate the outputs for values slightly more and slightly less than \(-3\) helps in predicting the trend in behavior.

Analyzing the function this way provides a clearer picture of how it operates—as the values approach a critical point. This method is invaluable when dealing with complex functions and helps in predicting and explaining function behavior at specific intervals. Being equipped with this knowledge improves problem-solving abilities in calculus and enhances understanding of different types of functions.