Math Problem Statement

Enter a T or an F in each answer space below to indicate whether the corresponding statement is true or false. You must get all of the answers correct to receive credit.

Solution

Let's analyze each statement and determine whether it's true (T) or false (F).

  1. Statement: "If p(x)p(x) is a polynomial, then the limit as x4x \to 4 of p(x)p(x) is p(4)p(4)."

    • Explanation: For any polynomial p(x)p(x), the limit as xx approaches a specific point aa is simply the value of p(x)p(x) at that point, p(a)p(a). This follows from the fact that polynomials are continuous functions.
    • Answer: True (T)
  2. Statement: limx3x2+4x20x2+5x22=limx3x2+4x20x2+5x22\lim_{x \to 3} \frac{x^2 + 4x - 20}{x^2 + 5x - 22} = \lim_{x \to 3} \frac{x^2 + 4x - 20}{x^2 + 5x - 22}

    • Explanation: First, let's simplify x2+4x20x2+5x22\frac{x^2 + 4x - 20}{x^2 + 5x - 22} as x3x \to 3.

      • Factor x2+4x20=(x+6)(x2)x^2 + 4x - 20 = (x + 6)(x - 2).
      • Factor x2+5x22=(x+6)(x3)x^2 + 5x - 22 = (x + 6)(x - 3).

      The expression simplifies to (x+6)(x2)(x+6)(x3)=x2x3\frac{(x + 6)(x - 2)}{(x + 6)(x - 3)} = \frac{x - 2}{x - 3}. As x3x \to 3, this simplifies to a form that does not exist (it leads to division by zero).

    • Answer: False (F)

  3. Statement: If limx3f(x)=4\lim_{x \to 3} f(x) = 4 and limx3g(x)=0\lim_{x \to 3} g(x) = 0, then limx3f(x)g(x)\lim_{x \to 3} \frac{f(x)}{g(x)} does not exist.

    • Explanation: Since g(x)g(x) approaches zero and f(x)f(x) approaches a non-zero value (4), the limit f(x)g(x)\frac{f(x)}{g(x)} will tend towards infinity or negative infinity, which means the limit does not exist in the finite sense.
    • Answer: True (T)
  4. Statement: If f(x)f(x) is differentiable at aa, then f(x)f(x) is continuous at aa.

    • Explanation: Differentiability at a point aa implies continuity at that point. If f(x)f(x) is differentiable at aa, it must also be continuous at aa.
    • Answer: True (T)
  5. Statement: If limx3f(x)=0\lim_{x \to 3} f(x) = 0 and limx3g(x)=0\lim_{x \to 3} g(x) = 0, then limx3f(x)g(x)\lim_{x \to 3} \frac{f(x)}{g(x)} does not exist.

    • Explanation: This is a case of an indeterminate form 00\frac{0}{0}. The existence of the limit depends on the behavior of f(x)f(x) and g(x)g(x) as xx approaches 3. For example, if both functions have the same linear behavior near x=3x = 3, the limit might exist. Therefore, this statement is not always true.
    • Answer: False (F)

Summary of Answers:

  1. True (T)
  2. False (F)
  3. True (T)
  4. True (T)
  5. False (F)

Would you like more details on any of these answers, or do you have questions?


Related Questions:

  1. What does it mean for a function to be continuous at a point?
  2. How does differentiability imply continuity?
  3. What are indeterminate forms, and how are they resolved?
  4. How can limits involving polynomials be simplified?
  5. What are some examples where f(x)g(x)\frac{f(x)}{g(x)} does exist even if f(x)0f(x) \to 0 and g(x)0g(x) \to 0?

Tip:

In limit problems involving rational functions, factoring and canceling common terms often help simplify the expression and resolve indeterminate forms.

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Math Problem Analysis

Mathematical Concepts

Limits
Polynomials
Continuity
Differentiability
Indeterminate Forms

Formulas

Limit properties
Factoring of polynomials

Theorems

Limit existence in rational functions
Differentiability implies continuity

Suitable Grade Level

Grade 11-12