Problem 1 - Precalculus: Mathematical for Calculus

Question

Suppose that \(\lim _{x \rightarrow a} f(x)=-3 \quad \lim _{x \rightarrow a} g(x)=0 \quad \lim _{x \rightarrow a} h(x)=8\) Find the value of the given limit. If the limit does not exist, explain why. (a) \(\lim _{x \rightarrow a}[f(x)+h(x)]\) (b) \(\lim _{x \rightarrow a}[f(x)]^{2}\) (c) \(\lim _{x \rightarrow a} \sqrt[3]{h(x)}\) (d) \(\lim _{x \rightarrow a} \frac{1}{f(x)}\) (e) \(\lim _{x \rightarrow a} \frac{f(x)}{h(x)}\) (f) \(\lim _{x \rightarrow a} \frac{g(x)}{f(x)}\) (g) \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}\) (h) \(\lim _{x \rightarrow a} \frac{2 f(x)}{h(x)-f(x)}\)

Step-by-Step Solution

Answer

(a) 5, (b) 9, (c) 2, (d) -1/3, (e) -3/8, (f) 0, (g) Does not exist, (h) -6/11.

Step 1: Sum of Limits

To find \( \lim_{x \rightarrow a} [f(x)+h(x)] \), use the property of limits that states \( \lim_{x \rightarrow a} (f(x) + h(x)) = \lim_{x \rightarrow a} f(x) + \lim_{x \rightarrow a} h(x) \). Substitute the given limits: \(-3 + 8 = 5\). Thus, \( \lim_{x \rightarrow a} [f(x)+h(x)] = 5 \).

Step 2: Limit of a Function Squared

The limit \( \lim_{x \rightarrow a} [f(x)]^{2} \) can be found by using the limit property \( \lim_{x \rightarrow a} [f(x)]^2 = (\lim_{x \rightarrow a} f(x))^2 \). Substitute the given limit: \((-3)^2 = 9\). Thus, \( \lim_{x \rightarrow a} [f(x)]^{2} = 9 \).

Step 3: Limit of a Cube Root

Use the property \( \lim_{x \rightarrow a} \sqrt[3]{h(x)} = \sqrt[3]{\lim_{x \rightarrow a} h(x)} \). Substitute the given limit for \( h(x) \): \( \sqrt[3]{8} = 2 \). Thus, \( \lim_{x \rightarrow a} \sqrt[3]{h(x)} = 2 \).

Step 4: Limit of the Reciprocal

For \( \lim_{x \rightarrow a} \frac{1}{f(x)} \), we need to know \( \lim_{x \rightarrow a} f(x) eq 0 \). Since \( \lim_{x \rightarrow a} f(x) = -3 \), substitute \( \frac{1}{-3} = -\frac{1}{3} \). Thus, \( \lim_{x \rightarrow a} \frac{1}{f(x)} = -\frac{1}{3} \).

Step 5: Limit of a Fraction

To find \( \lim_{x \rightarrow a} \frac{f(x)}{h(x)} \), use the limit property for division. The given limits are \( \lim_{x \rightarrow a} f(x) = -3 \) and \( \lim_{x \rightarrow a} h(x) = 8 \), so \( \frac{-3}{8} = -\frac{3}{8} \). Thus, \( \lim_{x \rightarrow a} \frac{f(x)}{h(x)} = -\frac{3}{8} \).

Step 6: Limit of a Zero Numerator

For \( \lim_{x \rightarrow a} \frac{g(x)}{f(x)} \), use \( \lim_{x \rightarrow a} g(x) = 0 \) and \( \lim_{x \rightarrow a} f(x) = -3 \). Substituting, \( \frac{0}{-3} = 0 \). Thus, \( \lim_{x \rightarrow a} \frac{g(x)}{f(x)} = 0 \).

Step 7: Uncertain Fraction

The expression \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \) involves dividing by zero \( \lim_{x \rightarrow a} g(x) = 0 \). In general, division by zero means the limit does not exist since \( f(x) eq 0 \). Thus, \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \) does not exist.

Step 8: Combining Limits with Adjustments

For \( \lim_{x \rightarrow a} \frac{2f(x)}{h(x) - f(x)} \), first evaluate the denominator \( h(x) - f(x) = 8 - (-3) = 11 \). Therefore, \( \frac{2(-3)}{11} = -\frac{6}{11} \). Conclusively, \( \lim_{x \rightarrow a} \frac{2f(x)}{h(x)-f(x)} = -\frac{6}{11} \).

Key Concepts

Limit Properties

In calculus, limit properties are fundamental rules that allow us to evaluate limits of functions more conveniently. These properties enable us to simplify complex expressions and make calculations manageable. The basic properties include:

• Sum Property: If you have two functions, say \( f(x) \) and \( h(x) \), and you are supposed to find the limit of their sum as \( x \) approaches a particular value, you simply add their individual limits. For example, \( \lim_{x \rightarrow a} (f(x) + h(x)) = \lim_{x \rightarrow a} f(x) + \lim_{x \rightarrow a} h(x) \).
• Product Property: Similar to addition, the limit of a product is the product of the limits. Meaning, \( \lim_{x \rightarrow a} [f(x) \cdot g(x)] = \lim_{x \rightarrow a} f(x) \cdot \lim_{x \rightarrow a} g(x) \).
• Difference and Quotient Properties: Limits of differences and quotients follow similar rules, provided the denominator is not zero in case of quotients.

Understanding these properties is key to evaluating limits quickly and accurately.

Division by Zero

Division by zero is a concept in mathematics that often leads to undefined expressions. In calculus, encountering a limit problem where the denominator approaches zero must be handled with care.When assessing limits that result in division by zero, it's crucial to determine whether it is leading to an indeterminate form or a specific asymptote.

• If you have a function \( \frac{f(x)}{g(x)} \) and \( \lim_{x \rightarrow a} g(x) = 0 \) while \( \lim_{x \rightarrow a} f(x) eq 0 \), the limit does not exist in the traditional sense, because the result is either infinity, negative infinity, or an indeterminate form.
• For example, \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \text{DNE (does not exist), if } g(x) = 0 \).

Handling division by zero properly is crucial for problem-solving in calculus. Always check each component of the fraction to avoid erroneous conclusions.

Fractional Limits

Fractional limits refer to problems involving limits of quotients, where both the numerator and denominator are approaching certain values.Here’s how to effectively approach fractional limits:

• Direct Substitution: First, try to substitute the values directly. If the denominator isn't zero, directly compute the fraction using the limits of the numerator and the denominator.
• Standard Form: If direct substitution leads to quotients like \( \frac{0}{0} \), further analysis is needed, often involving algebraic manipulation or simplifications to resolve the indeterminate form.
• Evaluation of Non-Zero Denominator: Ensure the denominator is non-zero for limits to exist in a fractional context, like \( \lim_{x \rightarrow a} \frac{f(x)}{h(x)} = \frac{-3}{8} \) in the given example.

Understanding and evaluating fractional limits allows students to extend their problem-solving to more complex rational functions.

Function Operations

In calculus, function operations are often an integral part of evaluating limits, particularly when dealing with combined or modified functions.Operations on functions concerning limits include addition, subtraction, multiplication, division, and compositions. Let's discuss some key points:

• Addition/Subtraction: When adding or subtracting functions under limits, break them into simpler parts using basic limit properties.
• Multiplication: Similarly, for multiplication, use the product of limits property. Keep in mind possible zero outcomes for more complex evaluation.
• Root and Power Functions: When dealing with power functions or roots, apply limits into \( [f(x)]^n \) or \( \sqrt[n]{f(x)} \), as shown by the cube root of 8 yielding 2.
• Reciprocals and Ratios: When working with reciprocals or ratios, ensure the denominator isn’t zero, particularly when evaluating as seen in the \( \frac{1}{f(x)} \) example, where checking \( f(x) eq 0 \) ensures the limit exists.

Using function operations correctly simplifies complex limit problems, helping to break them into more manageable parts.