Problem 23 - Contemporary Precalculus

Question

Find the limit if it exists. If the limit does not exist, explain why. $$\lim _{x \rightarrow 1^{+}}(\sqrt{x-1}+3)$$

Step-by-Step Solution

Answer

Answer: The limit of the given function as \(x\) approaches \(1^+\) is 3.

Step 1: Identify the Limit

We are asked to find the following limit: $$\lim _{x \rightarrow 1^{+}}(\sqrt{x-1}+3)$$

Step 2: Check for Direct Substitution

Before applying the limit definition, let's see if direct substitution works. Plug \(x=1\) into the expression: $$\sqrt{1-1}+3=0+3=3$$ Since the expression is well-defined at \(x=1\), and there is no indeterminate form, we can conclude that:

Step 3: State the Limit

The limit of the given function as \(x\) approaches \(1^+\) is: $$\lim _{x \rightarrow 1^{+}}(\sqrt{x-1}+3) = 3$$

Key Concepts

Direct Substitution

Direct substitution is one of the simplest methods for finding limits. It involves directly replacing the variable in the function with the number it is approaching. When we can do this without encountering an undefined expression or an indeterminate form, it makes the limit calculation straightforward. For the exercise, we looked at the limit \( \lim _{x \rightarrow 1^{+}}(\sqrt{x-1}+3) \). By plugging \( x = 1 \) directly into \( \sqrt{x-1}+3 \), we get \( \sqrt{1-1}+3 = 0+3 = 3 \). Since we arrived at a defined number without any complications, direct substitution confirms that the limit is 3.

Direct substitution is applicable in cases where the function is continuous at the point you're evaluating. More on continuity will be discussed in a later section. It's important to first check this method because it is quick and often effective. If the function results in any indeterminate form like \( \frac{0}{0} \) or \( \infty - \infty \), then we must look for other techniques such as factoring, rationalizing, or using L'Hôpital's Rule.

Right-Hand Limit

The concept of the right-hand limit, denoted as \( \lim_{x \to a^+} f(x) \), means we approach the point \( a \) from the right (or from values greater than \( a \)). This is crucial when the function behaves differently on either side of a point, particularly at boundaries or points of discontinuity.

In our example, the limit \( \lim _{x \rightarrow 1^{+}}(\sqrt{x-1}+3) \) specifies that we are only interested in how the function behaves as \( x \) approaches 1 from the positive side (or from values slightly greater than 1). This directional approach ensures that we don't encounter any undefined mathematical operations that might arise from taking the limit from the wrong direction.

Considering the function \( \sqrt{x-1} \), it's clear that \( x-1 \) must be non-negative for the square root to be real. By approaching from the right, we ensure \( x-1 \geq 0 \), thus making the expression valid. This makes right-hand limits incredibly useful for functions with domain restrictions or where the function’s behavior differs on either side of a point.

Continuity of Functions

Continuity of a function at a point means the function is unbroken and there are no gaps, jumps, or undefined points at or around that location. In mathematical terms, a function \( f(x) \) is continuous at \( x = a \) if:

• The function \( f(a) \) is defined.
• The limit \( \lim_{x \to a} f(x) \) exists.
• The value of the function at that point equals the limit: \( f(a) = \lim_{x \to a} f(x) \).

In the context of the given limit problem \( \lim _{x \rightarrow 1^{+}}(\sqrt{x-1}+3) \), we see that the function is not continuous at \( x=1 \) from the left because \( \sqrt{x-1} \) is only real for \( x > 1 \). However, the function is well-behaved and continuous from the right due to the square root term having valid input.

Ensuring a function's continuity can simplify limit evaluation since if a function is continuous at a certain point, the limit as \( x \) approaches that point is simply the value of the function at that point. This ties back to direct substitution, which works hand-in-hand with continuity, providing a quick way to find limits wherever applicable.