Problem 18 - Precalculus: Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

The limit is 4.

Step 1: Apply the Direct Substitution Property

First, check whether you can use the direct substitution property by plugging in the value that the variable approaches directly into the expression. Here, we substitute \( u = -2 \) into \( \sqrt{u^{4}+3u+6} \).

Step 2: Substitute and Simplify

Calculate \( u^4+3u+6 \) by plugging in \( u = -2 \). We get:\[-2^4 + 3(-2) + 6 = 16 - 6 + 6 = 16\]. This simplifies to \( \sqrt{16} \).

Step 3: Find the Square Root

Calculate the square root of 16, which is \( 4 \). Thus, \( \sqrt{16} = 4 \).

Step 4: Conclude the Limit Evaluation

Since we could directly substitute \( u = -2 \) and simplify the expression without any undefined behavior or discontinuity, we conclude:The limit \( \lim_{u \to -2} \sqrt{u^{4}+3u+6} = 4 \).

Key Concepts

Direct Substitution Property

The Direct Substitution Property is a straightforward method for evaluating limits. This property states that if a function is continuous at a point, you can find the limit of the function as it approaches that point simply by substituting the point into the function.

In this exercise, the goal was to evaluate the limit as \( u \) approaches \(-2\). We used the direct substitution property by directly plugging \( u = -2 \) into the expression \( \sqrt{u^{4}+3u+6} \). Since our function was continuous at \(-2\), the substitution showed no undefined behavior or discontinuities, thereby validating the use of this property.

Limit Laws

Limit Laws are fundamental rules that enable us to simplify the evaluation of limits. These include various properties such as the sum law, difference law, and the product law, among others.

These laws help you break down complex expressions into simpler parts when computing limits. In the given problem, while evaluating \( \lim_{u \to -2} \sqrt{u^4+3u+6} \), the calculation \( u^4+3u+6 \) could be processed, involving the application of basic arithmetic operations allowed by limit laws before applying the square root.

• Sum Law: Helps to calculate limits of sums individually.
• Product Law: Facilitates the calculation of limits for products.
• Power Law: Assists with calculations involving powers.

By using these laws, we ensure that each part of the expression is handled correctly, simplifying the overall limit evaluation process.

Continuity in Limits

Continuity is a central concept in calculus and particularly important when evaluating limits. A function is considered continuous at a point if the following three conditions hold: The function is defined at the point, the limit exists at that point, and the limit matches the function's value at that point.

In this exercise, the expression \( \sqrt{u^{4}+3u+6} \) entails operations which are continuous at \( u = -2 \). Since the expression involves basic polynomial and square root operations, which are known to be continuous where defined, the continuity assures that the direct substitution method will yield the correct limit. This continuity ensures there are no breaks or jumps where the limit fails to match the function value, making the computation much simpler and more reliable.

Square Root Function

The Square Root Function is crucial in many calculus problems. When dealing with limits, it's important to remember that the square root function, \( \sqrt{x} \), is continuous and defined for all non-negative values of \( x \).

In the problem \( \lim_{u \to -2} \sqrt{u^{4}+3u+6} \), the expression inside the square root, \( u^{4}+3u+6 \), evaluates to 16 when \( u = -2 \). Since 16 is a non-negative number, the square root of 16 is easily found.

Thus, we conclude with confidence that \( \sqrt{16} = 4 \), which makes the use of the square root valid in the computation of the limit. Understanding the domain and properties of the square root function ensures that we handle limits involving square roots accurately and effectively.