Problem 38 - Calculus: Early Transcendentals

Question

If \( 2x \le g(x) \le x^4 - x^2 +2 \) for all \( x \), evaluate \( \displaystyle \lim_{x \to 1}g(x) \).

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Identify bounds for g(x)

We are given that \( 2x \le g(x) \le x^4 - x^2 + 2 \). This inequality will help us apply the Squeeze Theorem to find the limit of \( g(x) \) as \( x \to 1 \).

Step 2: Find the limit of the lower bound

Consider the lower bound \( 2x \). Calculate \( \lim_{x \to 1} 2x \): \[\lim_{x \to 1}2x = 2\times 1 = 2.\]

Step 3: Find the limit of the upper bound

Consider the upper bound \( x^4 - x^2 + 2 \). Calculate \( \lim_{x \to 1} (x^4 - x^2 + 2) \): \[\lim_{x \to 1} (x^4 - x^2 + 2) = 1^4 - 1^2 + 2 = 2.\]

Step 4: Apply the Squeeze Theorem

Since we have found that both the lower bound \( \lim_{x \to 1} 2x \) and upper bound \( \lim_{x \to 1} (x^4 - x^2 + 2) \) equal 2, by the Squeeze Theorem, the limit of \( g(x) \) as \( x \to 1 \) is also 2.

Key Concepts

Limits

Limits are a fundamental concept in calculus, helping us understand the behavior of functions as they approach a certain point. Essentially, when we talk about the limit of a function as it approaches a specific value, we are interested in what value the function tends to get closer to.
For the given exercise, we want to find the limit of the function \( g(x) \) as \( x \to 1 \). This requires evaluating the behavior of \( g(x) \) as \( x \) gets indefinitely close to 1, but not necessarily when \( x \) is exactly 1.

Useful tips when dealing with limits:

• Check the behavior of functions as they approach the point from both left and right.
• If possible, try substituting the value directly into the function to see if it provides a meaningful result. This is especially helpful with polynomials.
• Watch for indeterminate forms, such as \( \frac{0}{0} \), which may require further manipulation or different techniques to resolve.

By leveraging these ideas, we can successfully address problems involving limits, such as the one in this exercise.

Inequalities

Inequalities provide boundaries within which the values of a function can lie. In this exercise, we know that \( g(x) \) is squeezed between two functions: \( 2x \) and \( x^4 - x^2 + 2 \). This establishes a useful constraint that allows us to analyze the behavior of \( g(x) \) without knowing its exact formula.
Understanding inequalities involves:

• Recognizing the importance of verifying that different parts of the inequality hold true across the required domain.
• Using inequalities to find limits can be particularly insightful when the exact definition of a function is unknown. They often appear in the Squeeze Theorem, which is a handy tool for determining limits.
• Simplifying inequalities can help set clearer boundaries, making it easier to apply limit laws effectively.

In this problem, applying inequalities effectively showed that \( g(x) \) approached the same value as its boundaries, confirming the result through the Squeeze Theorem.

Continuous Functions

Continuous functions are smooth and unbroken, providing a predictable manner of behavior across their domain. A function \( f(x) \) is continuous at \( x = a \) if \( \lim_{x \to a} f(x) = f(a) \). This implies that for continuous functions, approaching any point results in function values approaching a specific and unique number.
The significance of continuous functions in this context includes:

• Both the lower bound \( 2x \) and the upper bound \( x^4 - x^2 + 2 \) are continuous, which aids in evaluating limits straightforwardly.
• If the bounds of the inequalities are continuous functions, their limits can often be evaluated directly by substitution, simplifying the analysis significantly.
• Continuous functions reduce the complexity of applying the Squeeze Theorem because they maintain consistent behavior near the points of interest.

This property helped us smoothly use the Squeeze Theorem to find the limit of \( g(x) \) in this exercise by confirming that \( g(x) \) harmonizes with its continuous bounds as \( x \to 1 \).