Problem 31 - Calculus and its applications

Question

Use the Limit Properties to find the following limits. If a limit does not exist, state that fact. $$ \lim _{x \rightarrow 5} \sqrt{x^{2}-16} $$

Step-by-Step Solution

Answer

The limit is 3.

Step 1: Substitute x with 5

First, substitute the value of 5 into the function inside the limit: \( \sqrt{x^2 - 16} \). This gives: \( \sqrt{5^2 - 16} = \sqrt{25 - 16} \).

Step 2: Simplify the Expression

Now, simplify the expression inside the square root: \( \sqrt{25 - 16} = \sqrt{9} \).

Step 3: Calculate the Square Root

Finally, find the square root of 9 to determine the limit: \( \sqrt{9} = 3 \).

Step 4: State the Limit

Since there are no undefined points or discontinuity at \( x = 5 \), the limit exists. Hence, \( \lim _{x \rightarrow 5} \sqrt{x^{2}-16} = 3 \).

Key Concepts

Limit Properties

When working with limits, limit properties are essential tools that allow us to evaluate limits in a structured way. These properties give us rules to simplify and evaluate complex expressions.

Key limit properties include:

• Sum/Difference Rule: The limit of a sum/difference is the sum/difference of the limits.
• Product Rule: The limit of a product is the product of the limits.
• Quotient Rule: The limit of a quotient is the quotient of the limits, provided the denominator is not zero.
• Constant Multiple Rule: A constant can be factored out of a limit expression.
• Power Rule: The limit of a power is the power of the limit.

In the exercise, applying the limit directly by substitution is possible because the function under the limit is continuous around the point of interest (5), and there are no interruptions like division by zero or square roots of negatives.

Continuity

Continuity in a function is a vital concept when discussing limits. A function is considered continuous at a point if there is no abrupt change in its value around that point.

Here's what makes a function continuous at a point:

• Defined Value: The function must be defined at that point.
• Limit Existence: The limit of the function as it approaches the point must exist.
• Limit Equals Function Value: The limit of the function as it approaches the point must be equal to the function's value at that point.

For the expression \( \sqrt{x^2 - 16} \) at \( x = 5 \), it is continuous because:

• The expression is defined (no negative inside the square root).
• The limit exists and equals the function value (both are 3).

A continuous function at a given point means you can safely substitute and evaluate limits without unexpected "jumps" or undefined behavior.

Square Root

Understanding the square root is essential in many limits problems, especially those involving radicals. The square root symbol \( \sqrt{} \) refers to a number which, when multiplied by itself, gives the original value.

When dealing with square roots in limits:

• Ensure that the expression inside the square root (the radicand) is non-negative as square roots of negative numbers are not real.
• Calculate by simplifying the radicand first, then find the square root.

In the exercise, substituting \( x = 5 \) gives us \( \sqrt{9} \) which is straightforward to simplify.

• Start with \( \sqrt{25 - 16} \)
• Simplify inside: 25 - 16 = 9
• The square root is 3: \( \sqrt{9} = 3 \)

This straightforward approach emphasizes the fundamental rule that the processes inside limits should not encounter negative radicands, assuring real-valued results.