Problem 20 - Calculus

Question

Find the limits if they exist. An \(\varepsilon-\delta\) test is not required. $$ \lim _{x \rightarrow 2} \frac{\sqrt{4-x}}{\sqrt{6+x}} $$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: Identify the Limit Expression

We are given the expression \( \lim _{x \rightarrow 2} \frac{\sqrt{4-x}}{\sqrt{6+x}} \). Our task is to find the limit of this expression as \(x\) approaches 2.

Step 2: Substitute Directly

Substitute \(x = 2\) directly into the expression: \( \frac{\sqrt{4-2}}{\sqrt{6+2}} = \frac{\sqrt{2}}{\sqrt{8}} \).

Step 3: Simplify the Expression

Since \( \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2} \), the expression becomes \( \frac{\sqrt{2}}{2\sqrt{2}} \).

Step 4: Cancel Common Factors

Notice that the common factor \( \sqrt{2} \) can be canceled from both the numerator and the denominator. Thus, the expression simplifies to \( \frac{1}{2} \).

Key Concepts

Limit Laws

Limits in calculus help us understand the behavior of functions as they approach a specific point. To find limits, we can use several rules called Limit Laws. These laws make the process easier and more straightforward. When dealing with functions, Limit Laws help us break down complex expressions into simpler parts. Here are some common Limit Laws:

• 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.

These laws allow us to manipulate and find limits in a structured way. They are particularly helpful when limits involve addition, subtraction, multiplication, or division of functions. By applying these laws, we can solve complex limit problems with greater ease and efficiency.

Direct Substitution Method

The Direct Substitution Method is a simple and effective way to find limits, especially when the function is continuous at the point being considered. This method involves replacing the variable in the limit expression with the value it's approaching. If the resulting expression is not undefined or a "0/0" form, this method is a quick way to evaluate limits.For example, suppose we want to find \[ \lim _{x \rightarrow a} f(x) = f(a) \]This method is useful:- When the function is simple, and substituting "a" does not cause division by zero.- When the function is defined and continuous at the point "a".In the original exercise, direct substitution was used to substitute \(x = 2\) in the expression \( \frac{\sqrt{4-x}}{\sqrt{6+x}} \), resulting in \( \frac{\sqrt{2}}{\sqrt{8}} \). This direct substitution simplifies the process, allowing us to quickly determine the limit without complex calculations.

Simplifying Radicals

Simplifying radicals is an essential skill in algebra and calculus. It involves expressing a radical in its simplest form, making calculations and further simplifications more manageable. Simplifying radicals is especially useful when dealing with limits, as it can help clear up complex expressions.In mathematics, a radical expression is one that contains a root, such as a square root or cube root. To simplify a radical:

• Break down the number under the root into its prime factors.
• Look for pairs of factors (because we are often dealing with square roots).
• Move any pairs of factors out from under the root.

For instance, in the step by step solution, the expression \( \sqrt{8} \) was simplified by recognizing it as \( \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} \), which then becomes \( 2\sqrt{2} \). By simplifying \( \sqrt{8} \), easier cancellation of common factors was possible, ultimately leading to the simple fraction \( \frac{1}{2} \). Simplifying radicals, therefore, makes complex expressions easier to handle and solve.