Problem 11 - Thomas Calculus

Question

Find the limits in Exercises \(11-18\) $$\lim _{x \rightarrow-0.5} \sqrt{\frac{x+2}{x+1}}$$

Step-by-Step Solution

Answer

The limit is \( \sqrt{3} \).

Step 1: Substitute the Limit Point

To find \( \lim _{x \rightarrow -0.5} \sqrt{\frac{x+2}{x+1}} \), begin by substituting \( x = -0.5 \) directly into the expression: \( \frac{(-0.5) + 2}{(-0.5) + 1} \). Simplifying gives \( \frac{1.5}{0.5} = 3 \).

Step 2: Calculate the Square Root

Take the square root of the result from Step 1. The expression becomes \( \sqrt{3} \).

Step 3: Confirm the Continuity

Check whether the function \( \sqrt{\frac{x+2}{x+1}} \) is continuous at \( x = -0.5 \). The function inside the square root, \( \frac{x+2}{x+1} \), is continuous at \( x = -0.5 \) since the denominator is not zero. Therefore, the original function is also continuous at this point.

Step 4: Conclude the Limit

Since the function is continuous at \( x = -0.5 \), the limit equals the value of the function at this point. Thus, \( \lim _{x \rightarrow -0.5} \sqrt{\frac{x+2}{x+1}} = \sqrt{3} \).

Key Concepts

Understanding Continuous Functions

Continuous functions are foundational in calculus, making it easy to evaluate limits directly by substitution. In simple terms, a function is continuous at a point if there is no interruption or jump at that point. This means you can draw the function's graph at this point without lifting your pencil.

For a function to be continuous at a certain point, three conditions must be met:

• The function is defined at the point.
• The limit of the function as it approaches the point exists.
• The value of the function at that point is equal to the limit of the function as it approaches the point.

In the given exercise, the function \( \sqrt{\frac{x+2}{x+1}} \) is continuous at \( x = -0.5 \) since the denominator of the expression inside the square root, which is \( x+1 \), does not become zero. This ensures a smooth and uninterrupted curve, allowing us to directly use substitution to find the limit.

Exploring Square Roots

Square roots are used to find a number which when multiplied by itself gives the original number. It's one of the radicals in mathematics that students encounter early on. When dealing with limits involving square roots, it's important to ensure the expression inside the root is non-negative, since square roots are not defined for negative numbers in real numbers.

For the expression \( \sqrt{\frac{x+2}{x+1}} \), we ensure that \( \frac{x+2}{x+1} > 0 \) when examining limits. In our example, after substituting \( x = -0.5 \), the result inside the square root was 3. This is a positive number, so taking the square root is straightforward. The solution then became \( \sqrt{3} \), a basic application of the square root property.

Mastering the Substitution Method

The substitution method in calculus simplifies the process of finding limits, especially with continuous functions. It involves replacing the variable with the value it's approaching, checking if the function is defined and continuous at that point.

Here's a quick walkthrough of how the substitution worked in the example. Initially, \( x \) is replaced directly with \( -0.5 \) in the expression \( \sqrt{\frac{x+2}{x+1}} \). This substitution led to \( \frac{1.5}{0.5} = 3 \). The square root was then taken, and since the result was valid and non-negative, this confirmed the limit.

Substitution is often the easiest method to find limits when dealing with continuous functions because it provides a quick check on whether the value of the function at a certain point can represent the limit. It streamlines problems potentially filled with complex algebra or unpredictable factors, allowing for a clear path to the solution.