Fancy algebra to find a limit and make a function continuous | Differential Calculus | Khan Academy
TLDRThe video script discusses the continuity of a function f at x=5. It explains that for f to be continuous at x=5, the limit of f(x) as x approaches 5 must equal f(5), which is defined as c. The script then navigates through a mathematical process to find the value of c. Initially, it encounters an indeterminate form of 0/0 when substituting x=5 directly into the function. To resolve this, the script uses algebraic techniques to rationalize the numerator and simplify the expression. Ultimately, by taking the limit as x approaches 5, the script finds that c equals 1/6, confirming the function's continuity at x=5.
Takeaways
- 📝 The function f(x) is defined as the square root of (x + 4 - 3) / (x - 5) for x ≠ 5, and f(5) = c.
- 🔍 To determine the value of c for continuity at x = 5, we use the definition of continuity, which states that the limit as x approaches 5 of f(x) must equal f(5).
- 💡 The initial attempt to substitute x = 5 results in an indeterminate form of 0/0, which suggests the need for a different approach to find the limit.
- 📚 L'Hopital's rule is mentioned as a potential tool to handle indeterminate forms, but in this case, algebraic manipulation is used instead.
- 🌟 By multiplying the numerator and denominator by the conjugate expression (square root of (x + 4) + 3), the radical in the numerator can be eliminated.
- 🎯 After simplification, the expression becomes (x + 4 - 9) / (x - 5) * (sqrt(x + 4) + 3) / (sqrt(x + 4) + 3), which simplifies to 1 / sqrt(x + 4) + 3.
- 📈 Taking the limit as x approaches 5, the expression simplifies to 1 / (sqrt(5 + 4) + 3), which equals 1 / (sqrt(9) + 3).
- 🔢 Substituting the value of 5 into the simplified expression, we get 1 / (3 + 3), which results in the value of c being 1/6.
- 🎓 The continuity of the function at x = 5 is confirmed by showing that the limit of the function as x approaches 5 equals the function's value at x = 5, which is c = 1/6.
- 🌐 This problem demonstrates the application of algebraic techniques to rationalize expressions and understand the behavior of functions at certain points.
- 📊 The concept of limits and continuity is fundamental in calculus and helps in understanding the behavior of functions, especially at points of interest.
Q & A
What is the definition of continuity in the context of the given function?
-In the context of the given function, continuity at x equals 5 means that the limit as x approaches 5 of f of x is equal to f of 5. This implies that the function's value at x=5 is the same as its limit from the left and right approaches to x=5.
What is the given function f(x)?
-The given function f(x) is defined as the square root of (x + 4 - 3) over (x - 5) when x is not equal to 5, and f(x) is equal to a constant c when x equals 5.
What is the value of the function f(x) when x equals 5?
-The value of the function f(x) when x equals 5 is given as the constant c.
What is the indeterminate form that arises when directly substituting x=5 in the function?
-When directly substituting x=5 in the function, an indeterminate form of 0/0 arises because the numerator (5 + 4 - 3) becomes 0 and the denominator (x - 5) also becomes 0.
What is the purpose of multiplying by the radical in the algebraic manipulation?
-The purpose of multiplying by the radical in the algebraic manipulation is to eliminate the square root from the numerator, which allows for further simplification and evaluation of the limit as x approaches 5.
What is L'Hopital's rule, and how does it relate to the indeterminate form?
-L'Hopital's rule is a mathematical tool that provides a method for finding the limits of indeterminate forms, such as 0/0 or ∞/∞, by taking the derivatives of the numerator and denominator. In this context, it could potentially be used to find the limit as x approaches 5 for the given function.
How does the algebraic manipulation help in finding the limit of the function?
-The algebraic manipulation helps in finding the limit of the function by simplifying the expression and allowing for the cancellation of terms, which ultimately leads to a form that can be evaluated directly at x=5.
What is the simplified form of the function after the algebraic manipulation?
-After the algebraic manipulation, the simplified form of the function for x not equal to 5 is 1 over the square root of (x + 4 + 3), or 1 over the square root of (x + 7).
What is the limit of the function as x approaches 5 after simplification?
-The limit of the function as x approaches 5 after simplification is 1 over the square root of (5 + 4 + 3), which simplifies to 1 over the square root of 9, and finally to 1/3.
What is the value of the constant c that ensures continuity at x equals 5?
-The value of the constant c that ensures continuity at x equals 5 is 1/3, as the limit of the function as x approaches 5 must equal the value of the function at x=5.
How does the concept of rationalizing numerators or denominators relate to this problem?
-The concept of rationalizing numerators or denominators is used in this problem to eliminate the square root from the numerator by multiplying by a conjugate expression. This technique is often used to simplify expressions involving radicals and is key to solving the problem of finding the limit and the value of c for continuity.
Outlines
📚 Continuity and Function Evaluation
This paragraph discusses the concept of continuity in a mathematical function. It introduces a function, f(x), defined as the square root of (x + 4 - 3) / (x - 5) for x ≠ 5 and f(5) = c. The goal is to find the value of c that makes f continuous at x = 5. The explanation involves understanding the definition of continuity, which requires the limit of f(x) as x approaches 5 to be equal to f(5). However, direct substitution leads to an indeterminate form of 0/0. The paragraph then introduces algebraic techniques to resolve this form, specifically by multiplying the numerator and denominator by a conjugate expression to eliminate the square root, and simplifies the function to a form that allows for the calculation of the limit as x approaches 5.
🔢 Limit Calculation and Continuity Conclusion
This paragraph concludes the process of finding the value of c that ensures the continuity of the function f(x) at x = 5. It continues from the previous paragraph's algebraic simplification and demonstrates the calculation of the limit of the function as x approaches 5. By simplifying the expression and canceling out the (x - 5) term in the numerator and denominator, the limit is found to be 1/6. Therefore, the value of c is determined to be 1/6, confirming that the function is continuous at x = 5.
Mindmap
Keywords
💡Function
💡Continuity
💡Limit
💡Square Root
💡Indeterminate Form
💡L'Hopital's Rule
💡Algebraic Manipulation
💡Rationalize
💡Difference of Squares
💡Substitution
💡Principal Root
Highlights
The function f(x) is defined as the square root of (x + 4 - 3) over (x - 5), with the exception that f(x) = c when x equals 5.
To determine the value of c for continuity at x=5, we must find the limit of f(x) as x approaches 5, based on the definition of continuity.
Substituting x=5 directly into the function results in an indeterminate form of 0/0, which necessitates further analysis.
L'Hopital's rule could be applied to find the limit when faced with an indeterminate form, but we can also use algebraic techniques to solve this problem.
Rationalizing the numerator by multiplying both the numerator and denominator by the conjugate expression (square root of (x + 4) + 3) simplifies the function.
After rationalization, the function simplifies to (x + 4 - 9) over (x - 5) times the square root of (x + 4 + 3), which allows us to cancel out the (x - 5) term.
The simplified function becomes 1 over the square root of (x + 4 + 3), which can be used to find the limit as x approaches 5.
By substituting x=5 into the simplified function, we find that the limit and thus the value of c is 1/6.
The continuity of the function f(x) at x=5 is confirmed by finding that the limit as x approaches 5 equals the function value at x=5, which is c=1/6.
The problem-solving approach demonstrated in the transcript involves a combination of direct substitution, identification of indeterminate forms, and algebraic simplification.
The transcript provides a clear example of how to handle a limit problem that initially appears to be intractable due to an indeterminate form.
The use of rationalization is highlighted as a key algebraic technique for simplifying expressions involving square roots.
The transcript emphasizes the importance of understanding the definition of continuity in the context of limits and how it applies to function values.
The solution process showcases the transition from a complex expression to a simpler form that allows for the calculation of a limit.
The final value of c=1/6 demonstrates the successful application of algebraic techniques in solving a limit problem.
The transcript serves as an educational resource for those learning about limits, continuity, and algebraic simplification techniques.
The step-by-step explanation ensures that the process is transparent and understandable, making it a valuable learning tool.
The transcript's methodical approach to solving the problem can help build a strong foundation in mathematical problem-solving skills.
Transcripts
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