Problem 6 - Calculus: Early Transcendentals

Question

Evaluate the limit and justify each step by indicating the appropriate Limit Law(s). \( \displaystyle \lim_{u \to -2}\sqrt{u^4 + 3u + 6} \)

Step-by-Step Solution

Answer

The limit is 4.

Step 1: Understanding the Function

The function inside the limit is given by \( f(u) = \sqrt{u^4 + 3u + 6} \). We need to determine the limit of this function as \( u \to -2 \).

Step 2: Substitution

Before applying any limit laws, substitute \( u = -2 \) into the function to see if it is possible to evaluate directly. The expression becomes \( \sqrt{(-2)^4 + 3(-2) + 6} \).

Step 3: Calculating the Expression Inside the Square Root

Calculate \( (-2)^4 + 3(-2) + 6 \). This is \( 16 - 6 + 6 = 16 \).

Step 4: Apply Direct Substitution

Since no undefined behavior such as division by zero occurs, evaluate \( \sqrt{16} \), which simplifies to \( 4 \).

Step 5: Conclusion: Confirm the Limit

Since the expression under the square root remained defined upon substitution, by the direct substitution property of continuous functions, the limit is \( 4 \).

Key Concepts

Limit Laws

Limit laws are essential tools in calculus, allowing us to simplify and evaluate limits with ease. These rules are derived from fundamental properties of limits. Some common limit laws include:

• Sum Law: The limit of a sum is the sum of the limits.
• Difference Law: The limit of a difference is the difference of the limits.
• Product Law: The limit of a product is the product of the limits.
• Quotient Law: The limit of a quotient is the quotient of the limits (provided the denominator is not zero).
• Power Law: The limit of a function raised to a power is the limit of the function raised to that power.

When evaluating limits, these laws help us break down complex expressions into simpler parts that can be individually assessed. For the given exercise, recognizing these laws ensures that we understand why it's usually safe to evaluate limits by substituting values into functions unless certain conditions (like division by zero) are involved.

Direct Substitution Property

The direct substitution property is one of the simplest and most powerful tools when working with limits. It states that if a function is continuous at a point, the limit of the function as it approaches that point can be found simply by plugging the point into the function. This property can save time and effort if applicable. In our example, the function is \( f(u) = \sqrt{u^4 + 3u + 6} \). To evaluate the limit as \( u \to -2 \), the first step is to check if we can use direct substitution. By substituting \( u = -2 \) into the function, the expression inside the square root becomes \( 16 - 6 + 6 = 16 \). Since the square root of 16 is well-defined, without any division by zero or negative roots, the direct substitution indicates that the limit is 4. Direct substitution is particularly useful because it often simplifies complex calculations, provided the function remains continuous at the point of interest.

Continuous Functions

Continuous functions play a vital role in calculus, allowing for smooth evaluation of limits. A function is said to be continuous at a point \( x = c \) if the following conditions are met:

• The function \( f(c) \) is defined.
• The limit \( \lim_{x \to c}f(x) \) exists.
• The limit equals the function value: \( \lim_{x \to c}f(x) = f(c) \).

With continuous functions, you can confidently apply the direct substitution property. In this exercise, the function \( f(u) = \sqrt{u^4 + 3u + 6} \) is continuous around \( u = -2 \) because the operations inside the square root (addition, powers, multiplication) are inherently continuous. This ensures that, by continuity, substituting \( u = -2 \) directly will yield the correct limit. Having an understanding of continuity not only supports the reasons behind direct substitution but also aids in grasping more complex calculus problems.