Problem 5 - A Graphical Approach to Precalculus with Limits

Question

Let \(\lim _{x \rightarrow 4} f(x)=16\) and \(\lim _{x \rightarrow 4} g(x)=8 .\) Use the limit rules to find each limit. Do not use a calculator. $$\lim _{x \rightarrow 4} \sqrt{f(x)}$$

Step-by-Step Solution

Answer

4

Step 1: Identify the Given Limit

You are given that \( \lim_{x \to 4} f(x) = 16 \). This means as \( x \) approaches 4, the function \( f(x) \) approaches 16.

Step 2: Use the Limit of Root Rule

The limit of the root of a function \( \sqrt{f(x)} \) as \( x \to a \) is equal to the square root of the limit of the function: \( \lim_{x \to a} \sqrt{f(x)} = \sqrt{\lim_{x \to a} f(x)} \). Apply this to the given limit, so \( \lim_{x \to 4} \sqrt{f(x)} = \sqrt{\lim_{x \to 4} f(x)} \).

Step 3: Substitute the Known Limit into the Formula

Substituting the known limit into the equation, you get \( \sqrt{\lim_{x \to 4} f(x)} = \sqrt{16} \).

Step 4: Calculate the Square Root

Calculate the square root of 16. \( \sqrt{16} = 4 \).

Step 5: Conclude the Result

The limit of \( \sqrt{f(x)} \) as \( x \to 4 \) is 4. This follows from Step 4 where the square root of 16 was calculated to be 4.

Key Concepts

Limit Rules

Calculating limits is a crucial part of understanding calculus. Let's explore the fundamental limit rules. When approaching a limit, you are essentially finding out what value a function approaches as the variable gets closer and closer to a certain point.
For our exercise, we needed to ascertain the limit of a square root function, specifically \( \lim_{x \to 4} \sqrt{f(x)} \). Here, we employ the Limit of Root Rule. This rule is highly useful whenever you need to find the limit of a function involving roots or powers.

• The Limit of a Constant Rule: If \( c \) is a constant, then \( \lim_{x \to a} c = c \).
• The Limit of a Function Rule: For any function \( f(x) \), \( \lim_{x \to a} f(x) = L \) implies the function approaches \( L \) as \( x \) approaches \( a \).
• The Limit of Sum and Difference: \( \lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) \).
• The Limit of a Product: \( \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \), provided both limits exist.
• The Limit of a Quotient: \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \), given \( \lim_{x \to a} g(x) eq 0 \).

These rules help us determine how functions behave as they approach specific points. This knowledge allows us to tackle even more complex functions confidently.

Square Root Function

The square root function is a common mathematical function that often appears in limits and other expressions. Understanding its properties is important for solving limit problems.
The square root of a number is a value that, when multiplied by itself, gives the original number. It's denoted as \( \sqrt{x} \). For example, \( \sqrt{16} \) is 4 because \( 4 \times 4 = 16 \). This simplicity allows us to convert complex expressions like \( \sqrt{f(x)} \) into easier components when calculating limits.
The Limit of Root Rule, as used in our exercise, demonstrates how we can directly apply the limit to the function under the root:
\[ \lim_{x \to a} \sqrt{f(x)} = \sqrt{\lim_{x \to a} f(x)} \]
This means you can take the limit first and then calculate the square root, simplifying the process. This rule is straightforward but requires the understanding that the inner function \( f(x) \) must have a limit at \( x \) approaching "a". It also relies on the fact that the square root function is continuous for non-negative numbers, which assures the existence of such limits.

Substitution Method

The Substitution Method is a straightforward approach to solving limits, especially useful when direct evaluation leads to indeterminate forms. In simple cases like our exercise, where you already know \lim_{x \to 4} f(x) = 16\, substitution becomes a powerful ally to streamline the process.
When using the substitution method, the key steps are:

• Identify when substitution is applicable. This typically involves recognizing that the function inside the limit approaches a known value.
• "Substitute" the known limit directly into the function when possible. This allows the evaluation to become much simpler.
• Calculate the simpler expression, such as finding the square root of a constant after substitution.

In our exercise, we substituted \( \lim_{x \to 4} f(x) = 16 \) to find the expression \( \sqrt{\lim_{x \to 4} f(x)} \). This direct substitution yielded \( \sqrt{16} = 4 \), swiftly providing the limit result without additional computation.
The elegance of substitution lies in its simplicity, reducing complex problems into basic arithmetic operations once the conditions are favorable, aiding both comprehension and calculation.