Question

One-sided limits a. Evaluate \(\lim _{x \rightarrow 2^{+}} \sqrt{x-2}\). b. Explain why \(\lim _{x \rightarrow 2^{-}} \sqrt{x-2}\) does not exist.

Short answer

Answer: The limit as x approaches 2 from the right is 0, and the limit as x approaches 2 from the left does not exist.

Step-by-step solution

Understand the function and its behavior around x=2

We are given the function \(f(x) = \sqrt{x-2}\). As x approaches 2 from either side, the expression inside the square root gets closer and closer to 0. When x is greater than 2 (i.e., x approaches 2 from the right), the square root will be computed on a positive number giving a real value. However, when x is less than 2 (i.e., x approaches 2 from the left), the square root will be computed on a negative number, which results in a non-real value (an imaginary number). This information will be useful in evaluating the one-sided limits.

Evaluate the limit as x approaches 2 from the right

To evaluate the limit as x approaches 2 from the right (\(x \rightarrow 2^{+}\)), we can simply substitute x with a value slightly greater than 2 in the function and see how the function behaves: $$\lim _{x \rightarrow 2^{+}} \sqrt{x-2} = \sqrt{2.001 - 2} = \sqrt{0.001}$$ Since this results in a real number just above 0, the limit as x approaches 2 from the right is: $$\lim _{x \rightarrow 2^{+}} \sqrt{x-2} = 0$$

Explain why the limit as x approaches 2 from the left does not exist

As we observed in Step 1, when x approaches 2 from the left (\(x \rightarrow 2^{-}\)), the expression inside the square root gets closer to 0 but remains negative. That means we are trying to evaluate the square root of a negative number, which yields a non-real (imaginary) result, like so: $$\lim _{x \rightarrow 2^{-}} \sqrt{x-2} = \sqrt{1.999 - 2} = \sqrt{-0.001}$$ This results in an imaginary number, which indicates that the limit as x approaches 2 from the left does not exist. This is because a limit can only be considered meaningful if it results in a real number. Thus, we can conclude that: $$\lim _{x \rightarrow 2^{-}} \sqrt{x-2}\text{ does not exist.}$$

Key concepts

Understanding Limits

Limits help us understand the behavior of a function as it approaches a particular point. In mathematics, a limit describes how a function behaves as its input gets closer to a certain value. For example, when we say \( \lim_{x \to a} f(x) \), we're interested in what value \( f(x) \) is getting closer to as \( x \) gets closer to \( a \).

**One-sided Limits**
Sometimes, we look at limits from just one direction:

• \( \lim_{x \to a^+} f(x) \): Approaches from the right.
• \( \lim_{x \to a^-} f(x) \): Approaches from the left.

In our exercise, we're examining the function \( f(x) = \sqrt{x-2} \) as \( x \) approaches 2 from both directions.

Exploring Real Numbers

Real numbers include all the numbers on the continuous number line:

• Natural numbers (1, 2, 3, ...)
• Whole numbers (0, 1, 2, ...)
• Integers (-1, 0, 1, ...)
• Rational numbers (fractions like 1/2, 3/4)
• Irrational numbers (non-repeating, non-terminating decimals like \( \pi \), \( \sqrt{2} \))

A real number is the result when a function is approaching a specific limit, making it meaningful and significant. In the problem given, the function \( \sqrt{x-2} \) has a real number limit when \( x \) approaches 2 from the right (\( 0 \)), but not when approaching from the left.

Imaginary Numbers Explained

Imaginary numbers come into play when dealing with square roots of negative numbers. They help us solve equations that don't have real solutions. The basic imaginary unit is \( i \), where \( i^2 = -1 \).

When the exercise examined \( \sqrt{x-2} \) as \( x \) approached 2 from the left, the expression inside the square root became negative, e.g., \( \sqrt{-0.001} \). This results in an imaginary number, like \( 0.0316i \), indicating the limit does not exist in the real number realm.

Diving Into Square Roots

A square root finds the number that, when multiplied by itself, equals the given number. For example, \( \sqrt{4} = 2 \) because \( 2 \times 2 = 4 \).

Square roots of positive numbers lead to real results, while square roots of negative numbers lead to imaginary numbers. In the previous example, when \( x \to 2^+ \), \( \sqrt{x-2} \) approaches different values on the number line:

• \( x = 2^+ \): \( \sqrt{x-2} \) yields a positive square root (real number).
• \( x = 2^- \): \( \sqrt{x-2} \) yields a negative square root (imaginary number).

This difference in square root results explains the existence and non-existence of one-sided limits.