Problem 64 - Thomas Calculus

Question

If $$2-x^{2} \leq g(x) \leq 2 \cos x \text { for all } x, \text { find } \lim _{x \rightarrow 0} g(x).$$

Step-by-Step Solution

Answer

The limit \( \lim_{x \to 0} g(x) = 2 \).

Step 1: Understand the Problem

We are given that the function \( g(x) \) is bounded by two functions: \( 2-x^2 \) and \( 2 \cos x \). We need to find the limit of \( g(x) \) as \( x \) approaches 0.

Step 2: Find the Left Bound Limit

Analyze the limit of the lower bound as \( x \to 0 \). For the function \( 2 - x^2 \), as \( x \to 0 \), \( x^2 \to 0 \), so \( 2 - x^2 \) approaches 2. Thus, \( \lim_{x \to 0} (2 - x^2) = 2 \).

Step 3: Find the Right Bound Limit

Analyze the limit of the upper bound as \( x \to 0 \). For the function \( 2 \cos x \), as \( x \to 0 \), \( \cos x \to 1 \), so \( 2 \cos x \) approaches 2. Therefore, \( \lim_{x \to 0} (2 \cos x) = 2 \).

Step 4: Apply Squeeze Theorem

Since \( 2 - x^2 \leq g(x) \leq 2 \cos x \) for all \( x \), and both bounding functions approach 2 as \( x \to 0 \), the Squeeze Theorem implies that \( \lim_{x \to 0} g(x) = 2 \).

Key Concepts

Limit of a Function

Understanding the concept of the limit is vital in calculus. Simply put, a limit helps us determine what value a function approaches as the input (or 'x') gets closer to a specific point. In mathematical terms, we say the limit of some function \( f(x) \) as \( x \) approaches \( c \) is \( L \) if we can get \( f(x) \) as close to \( L \) as we like by making \( x \) close enough to \( c \). We denote this as \( \lim_{x \to c} f(x) = L \).

For example, in our exercise, we are interested in finding \( \lim_{x \to 0} g(x) \). To find this limit, we look at how \( g(x) \) behaves as \( x \) tends to zero. By examining the behavior of \( g(x) \) between its bounding functions, we use the properties of limits to deduce the solution. This is a powerful idea because it helps us deal with functions that are complicated or not straightforward to evaluate directly, such as those involving oscillating or undefined terms at a point.

Bounding Functions

Bounding functions play a crucial role in understanding the behavior of complex functions. Essentially, these are functions which stay above and below another function within a certain interval, thereby "bounding" it. In calculus, bounding functions are instrumental in evaluating limits, especially when direct computation is difficult.

In our case, the function \( g(x) \) is squeezed between two bounding functions: \( 2 - x^2 \) and \( 2 \cos x \). Here, for all \( x \), the inequality \( 2 - x^2 \leq g(x) \leq 2 \cos x \) holds. This tells us that no matter the behavior of \( g(x) \) precisely, it cannot exceed the behavior of \( 2 \cos x \) and will not be less than \( 2 - x^2 \).

Bounding functions help establish limits through the Squeeze Theorem because they create an interval within which the target function must exist, making it easier to predict the approach of its limits.

Calculus Problem Solving

Problem-solving in calculus often involves a mix of conceptual understanding and procedural techniques. Calculus challenges, like finding limits, require you to systematically address functions through a series of logical steps. One powerful method is to evaluate limits using concepts like bounding functions and the Squeeze Theorem.

When faced with the original problem, the steps involve:

• First, identify the bounding functions and confirm their limits as \( x \rightarrow 0 \).
• Next, apply the Squeeze Theorem by recognizing that the limit of \( g(x) \) must match the limits of the bounding functions at 0.
• Finally, conclude that \( \,\lim_{x \to 0} g(x) = 2 \) because the bounding limits equate to the same value.

By breaking down problems into these digestible steps and thoroughly understanding the concepts of limits and bounding functions, calculus problems become more manageable and solvable. This structured approach is key to mastering calculus.