Problem 44 - A Graphical Approach to Precalculus with Limits

Question

Determine each limit, if it exists. $$\lim _{x \rightarrow 1} \sqrt{3-x}$$

Step-by-Step Solution

Answer

The limit is \( \sqrt{2} \).

Step 1: Substitute the Limit Value

To evaluate the limit \( \lim _{x \rightarrow 1} \sqrt{3-x} \), start by directly substituting \( x = 1 \) into the expression \( \sqrt{3-x} \). Substitute to get \( \sqrt{3-1} = \sqrt{2} \).

Step 2: Simplify the Expression

The expression simplifies to \( \sqrt{2} \) because \(3 - 1 = 2\). The square root of 2 is an irrational number, but it is a valid real number value.

Step 3: Determine Existence of the Limit

Since we have substituted and obtained a real number \( \sqrt{2} \), this means the limit exists. The continuity of \( \sqrt{3-x} \) at \( x = 1 \) ensures the limit is valid.

Key Concepts

Substitution Method

The substitution method is an essential tool in calculus for evaluating limits. It involves directly replacing the variable in a function with a specific value to simplify the calculation of a limit. Essentially, you "plug in" the limit value.

• Substitute the variable with the value it's approaching.
• Calculate the resulting expression to see if it simplifies to a real number.
• If the expression becomes undefined or doesn't resolve to a real number, other methods may be needed, such as factoring or rationalizing.

In our exercise, to find the limit \( \lim_{x \rightarrow 1} \sqrt{3-x} \), we substitute \( x \) with 1. This simplification leads us directly to the expression \( \sqrt{2} \). Since this is a real number, substitution worked perfectly, confirming that the limit exists.

Continuity

Continuity is a crucial principle when considering limits. A function is continuous at a point if, as the variable approaches that point, the function value approaches the limit without interruption. This can be tested using the substitution method.
For a function \( f(x) \) to be continuous at a point \( x = c \), the following conditions must be met:

• The function \( f(x) \) is defined at \( x = c \).
• The limit \( \lim_{x \rightarrow c} f(x) \) exists.
• The limit equals the function value, \( f(c) \).

In our example, the function \( \sqrt{3-x} \) is continuous at \( x = 1 \) because we can substitute directly to find that the limit exists and equals \( \sqrt{2} \). The continuity here ensures that the behavior of the function does not involve any jumps, breaks, or oscillations at that point.

Irrational Numbers

Irrational numbers are numbers that cannot be expressed as a simple fraction. This means their decimal form is non-terminating and non-repeating. Despite being "irrational," they are a fundamental part of the real number system and show up often in calculus.

Common examples include \( \sqrt{2} \), \( \pi \), and \( e \).

They serve as solutions to various equations and appear in many limits and geometric contexts.

In the context of our limit problem, when we computed \( \sqrt{3-x} \) as \( \sqrt{2} \), we found an irrational number as the result. This does not affect the existence of the limit but highlights how limits can converge to these non-fractional real numbers seamlessly. This demonstrates that irrational numbers are values where limits can and do exist routinely.