Problem 89 - Calculus

Question

Suppose that \(\lim _{x \rightarrow 3} f(x)=7 .\) Are the statements in Problems \(89-95\) true or false? If a statement is true, explain how you know. If a statement is false, give a counterexample. $$\lim _{x \rightarrow 3}(x f(x))=21$$

Step-by-Step Solution

Answer

True, because \( \lim_{x \to 3}(x) \cdot \lim_{x \to 3}(f(x)) = 21 \).

Step 1: Understanding the Limit Rule

When evaluating limits of products, the rule states that \( \lim_{x \to a} (g(x) \cdot h(x)) = \lim_{x \to a} g(x) \cdot \lim_{x \to a} h(x) \) provided both limits exist. Here, we need to identify \( g(x) \) and \( h(x) \).

Step 2: Identify the Functions

Recognize that in the expression \( \lim_{x \to 3}(x f(x)) \), we have \( g(x) = x \) and \( h(x) = f(x) \). The problem statement provides \( \lim_{x \to 3} f(x) = 7 \).

Step 3: Calculate Limit of x

Calculate \( \lim_{x \to 3} x \). Since \( x \) is a linear and continuous function, \( \lim_{x \to 3} x = 3 \).

Step 4: Apply the Limit Product Rule

Apply the limit product rule: \[ \lim_{x \to 3} (x f(x)) = \lim_{x \to 3} x \cdot \lim_{x \to 3} f(x) = 3 \cdot 7 = 21 \].

Step 5: Conclusion

Since both limits exist and the calculation confirms \( 3 \cdot 7 = 21 \), the statement \( \lim_{x \to 3}(x f(x))=21 \) is true.

Key Concepts

Limit Product Rule

When dealing with the limit of a product of two functions, the Limit Product Rule becomes essential. This rule states that if you have two functions \( g(x) \) and \( h(x) \), and both limits exist as \( x \to a \), then \( \lim_{x \to a} (g(x) \cdot h(x)) = \lim_{x \to a} g(x) \cdot \lim_{x \to a} h(x) \). This means that you can find the limit of a product by finding the individual limits first, and then multiplying them together. This rule simplifies a big problem by breaking it into smaller, manageable parts.

For example, in the given problem, we have \( g(x) = x \) and \( h(x) = f(x) \). We know \( \lim_{x \to 3} f(x) = 7 \). Therefore, \( \lim_{x \to 3} (x f(x)) \) can be calculated by simply taking the limit of \( g(x) \) as \( x \to 3 \) which is 3, and multiplying it by the limit of \( h(x) \) or \( f(x) \), which is 7. Hence, \( 3 \cdot 7 = 21 \).

Using the Limit Product Rule is particularly helpful when dealing with complex expressions that would otherwise be challenging to handle directly. It just shows how powerful and efficient this mathematical tool can be.

Continuous Functions

Understanding the concept of continuous functions is crucial while evaluating limits. A function is continuous at a point if there is no interruption or jump at that point. More formally, a function \( f(x) \) is said to be continuous at \( x = a \) if the limit as \( x \to a \) is exactly equal to the function's value at \( x = a \) itself, i.e., \( \lim_{x \to a} f(x) = f(a) \).

Linear functions, like \( g(x) = x \) from the example, are classic examples of continuous functions. Why? Because they don't have any breaks or holes; they can be drawn without lifting the pencil from the paper.

This property makes continuous functions predictable and easier to handle when calculating limits. You can safely swap evaluating a limit with simply finding the value of the function at a specific point.

So, for finding \( \lim_{x \to 3} x \), where \( g(x) = x \) which is continuous, we can directly substitute \( x = 3 \) to get \( g(3) = 3 \). Simple, right?

Evaluating Limits

Evaluating limits involves finding what value a function approaches as the input approaches a certain point. This concept is vital in calculus because it deals with various questions about how functions behave near certain points, especially when dealing with points not included in the domain.

Here’s the process:

• Identify the point \( a \) to which \( x \) is approaching.
• Evaluate the limit by substitution if the function is continuous at that point.
• If direct substitution results in undefined terms, explore other strategies like factorization, rationalization, or special limit laws applicable.

For instance, to evaluate \( \lim_{x \to 3} (x f(x)) \), recognizing \( f(x) \)'s limit, and using the Limit Product Rule helps us compute this directly if both functions involved are continuous at \( x = 3 \).

In this case, the evaluation becomes straightforward because both outer product components are nicely behaved around the point x = 3, leading us to \( 21 \) effortlessly. Hence, having a firm grip on limit evaluation techniques is incredibly useful for solving limits accurately and efficiently.