Math Problem Statement
Solution
Let’s go through each statement one by one and analyze whether each is true or false.
-
If and , then does not exist.
- When the limit of the denominator is 0 while the limit of the numerator is non-zero (here, 6), the limit typically does not exist due to division by zero.
- Answer: True
-
If exists, then the limit is .
- The existence of implies that is continuous at , meaning .
- Answer: True
-
If is differentiable at , then is continuous at .
- Differentiability at a point implies continuity at that point. If is differentiable at , then must also be continuous at .
- Answer: True
-
If is a polynomial, then is .
- Polynomials are continuous everywhere. Therefore, .
- Answer: True
-
If is continuous at , then is differentiable at .
- Continuity at a point does not guarantee differentiability. A function can be continuous but not differentiable (e.g., at ).
- Answer: False
Summary of Answers:
- True
- True
- True
- True
- False
Would you like more details on any of these answers or have any other questions?
Related Questions to Expand Understanding:
- What are the conditions required for differentiability at a point?
- Can you provide an example of a function that is continuous but not differentiable at a certain point?
- How does the Squeeze Theorem relate to limits involving division by zero?
- Why does differentiability imply continuity, but not vice versa?
- What are some properties of polynomial functions regarding limits and continuity?
Tip:
When evaluating limits involving fractions, always check if the limit of the denominator is zero, as this can indicate a potential for non-existence or the need for further analysis, such as applying L'Hôpital's Rule.
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Math Problem Analysis
Mathematical Concepts
Limits
Continuity
Differentiability
Polynomials
Formulas
\lim_{x \to a} f(x) = L
f'(a)
f(x) is continuous at a
f(x) is differentiable at a
Theorems
Limit theorem
Continuity and differentiability
Polynomial continuity
Suitable Grade Level
Undergraduate Calculus
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