Problem 45 - Precalculus with Limits

Question

In Exercises 45 and 46, use the given information to evaluate each limit. $$\lim_{x \to c} f(x)=3, \quad \lim_{x \to c} g(x)=6$$ $$\textrm{(a)} \quad \lim_{x \to c}\ [-2g(x)] \quad \quad \quad \quad \quad \textrm{(b)} \lim_{x \to c}\ [f(x)+g(x)]$$ $$\textrm{(c)} \quad \lim_{x \to c}\ \dfrac{f(x)}{g(x)} \quad \quad \quad \quad \quad \textrm{(d)} \lim_{x \to c}\ \sqrt{f(x)}$$

Step-by-Step Solution

Answer

The solutions are -12 for (a), 9 for (b), 0.5 for (c), and \(\sqrt{3}\) for (d).

Step 1: Evaluate the limit (a)

To find the limit of \(-2g(x)\) as \(x\) approaches \(c\), multiply the limit of \(g(x)\) as \(x\) approaches \(c\) by -2. By using the rule that the limit of a constant times a function is the constant times the limit of the function, we get: \(-2 * \lim_{x \to c} g(x) = -2 * 6 = -12\)

Step 2: Evaluate the limit (b)

To find the limit of \(f(x) + g(x)\) as \(x\) approaches \(c\), add the limits of \(f(x)\) and \(g(x)\) as \(x\) approaches \(c\). By using the rule that the limit of a sum is the sum of the limits, we get: \(\lim_{x \to c} f(x) + \lim_{x \to c} g(x) = 3 + 6 = 9\)

Step 3: Evaluate the limit (c)

To find the limit of the quotient \(f(x)/g(x)\) as \(x\) approaches \(c\), divide the limit of \(f(x)\) as \(x\) approaches \(c\) by the limit of \(g(x)\) as \(x\) approaches \(c\). By using the rule that the limit of a quotient is the quotient of the limits (given the limit of the denominator isn't zero), we get: \(\lim_{x \to c} f(x) / \lim_{x \to c} g(x) = 3 / 6 = 0.5\)

Step 4: Evaluate the limit (d)

To find the limit of \(\sqrt{f(x)}\) as \(x\) approaches \(c\), take the square root of the limit of \(f(x)\) as \(x\) approaches \(c\). By using the rule that the limit of a square root of the function is the square root of the limit of the function, we get: \(\sqrt{\lim_{x \to c} f(x)} = \sqrt{3}\)

Key Concepts

Limit of a Function

In mathematics, understanding the "limit of a function" is crucial for analyzing how functions behave as their input values approach a certain point. Essentially, the limit of a function at a point provides us with the value that the function approaches as the input nears that point. It is not necessary that the function reaches this value; what's important is the behavior of the function as it gets closer to that point.
To illustrate, consider the function given by the exercise: \( \lim_{x \to c} f(x) = 3 \) and \( \lim_{x \to c} g(x) = 6 \). This notation indicates that as \( x \) gets infinitely close to \( c \), \( f(x) \) gets closer and closer to 3, and \( g(x) \) approaches 6. These "limit values" tell us what to expect from the functions without necessarily evaluating them at \( x = c \).
In calculus, limits help us to define other major concepts like derivatives and integrals, making them foundational building blocks. Therefore, mastering this concept paves the way for understanding more advanced topics in mathematics.

Limit Rules

The calculation of limits often involves applying a set of "limit rules". These rules simplify the process, allowing us to resolve limits without directly substituting values, which is especially useful with complex functions, or where direct substitution leads to undefined forms.
Here are some essential limit rules:

• Sum Rule: The limit of a sum is the sum of the limits. For example, \( \lim_{x \to c} [f(x) + g(x)] = \lim_{x \to c} f(x) + \lim_{x \to c} g(x) \), which the exercise showed as \(3 + 6 = 9\).
• Constant Multiple Rule: The limit of a constant multiplied by a function is the constant multiplied by the limit of the function. For instance, \(-2 \cdot \lim_{x \to c} g(x) = -12\).
• Quotient Rule: The limit of a quotient is the quotient of the limits—provided the limit of the denominator is not zero—as shown \(3/6 = 0.5\).
• Root Rule: The limit of a root of a function is the root of the limit of the function, demonstrated as \( \sqrt{3} \).

These rules are powerful tools for solving problems involving limits. Understanding how to apply them correctly is key to solving calculus problems efficiently.

Continuous Functions

A "continuous function" is one where small changes in the input result in small changes in the output. This means there are no abrupt jumps, breaks, or holes in the graph of the function. Continuity is an important property because if a function is continuous at a point, it means the limit at that point equals the function's value.
In the framework of limits, a function \( f(x) \) is continuous at \( x = c \) if:

• The limit \( \lim_{x \to c} f(x) \) exists.
• The function \( f(x) \) is defined at \( x = c \).
• The limit of the function equals the function's value at that point: \( \lim_{x \to c} f(x) = f(c) \).

This exercise does not directly state whether the functions are continuous, but generally, having a limit helps analyze continuity. If \( f(x) \) and \( g(x) \) are both continuous at \( x = c \), this can greatly simplify problem-solving because continuity ensures nice behavior at the point and around it. Understanding continuous functions is vital for deeper insights into calculus and ensures a smoother understanding of the flow of functions.