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Questions and Answers
Given $\lim_{x \to c} f(x) = L$, which of the following statements provides the most accurate interpretation of the epsilon-delta definition of a limit?
Given $\lim_{x \to c} f(x) = L$, which of the following statements provides the most accurate interpretation of the epsilon-delta definition of a limit?
- For every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - c| < \delta$, then $|f(x) - L| < \epsilon$ (correct)
- For every $\delta > 0$, there exists an $\epsilon > 0$ such that if $0 < |x - c| < \epsilon$, then $|f(x) - L| < \delta$.
- For every $\epsilon > 0$, there exists a $\delta > 0$ such that if $|x - c| < \delta$, then $|f(x) - L| < \epsilon$.
- For every $\epsilon > 0$, there exists a $\delta > 0$ such that if $|x - c| < \epsilon$, then $0 < |f(x) - L| < \delta$.
If $\lim_{x \to c} f(x)$ and $\lim_{x \to c} g(x)$ both exist, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)}$ is always true, regardless of the value of $\lim_{x \to c} g(x)$.
If $\lim_{x \to c} f(x)$ and $\lim_{x \to c} g(x)$ both exist, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)}$ is always true, regardless of the value of $\lim_{x \to c} g(x)$.
False (B)
Using the properties of limits, evaluate $\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$.
Using the properties of limits, evaluate $\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$.
12
According to the definition of limits, for $\lim_{x \to c} f(x) = L$, the value of $f(c)$ is ______ relative to the existence and value of the limit.
According to the definition of limits, for $\lim_{x \to c} f(x) = L$, the value of $f(c)$ is ______ relative to the existence and value of the limit.
Match the limit properties with their corresponding descriptions:
Match the limit properties with their corresponding descriptions:
Flashcards
What is a limit?
What is a limit?
The value that a function 'approaches' as the input 'approaches' some value.
Formal definition of a limit
Formal definition of a limit
If f(x) gets arbitrarily close to L for all x sufficiently close to c, then the limit of f(x) as x approaches c is L.
Limit of a constant
Limit of a constant
The limit of a constant 'b' as x approaches 'c' is simply 'b'.
Limit properties
Limit properties
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Limit of a power function
Limit of a power function
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Study Notes
Definition of Limits
- For a function f defined on an open interval containing c (except possibly at c), the limit of f(x) as x approaches c equals L which is written as lim (x→c) f(x) = L.
- For every ε>0, there exists a δ>0 such that if 0<|x-c|<δ, then |f(x)-L|<ε.
Basic Limits
- If b and c are real numbers and n is a positive integer, then:
- lim (x→c) b = b
- lim (x→c) x = c
- lim (x→c) x^n = c^n
Examples of Basic Limits
- lim (x→2) 3 = 3
- lim (x→4) x = -4
- lim (x→2) x² = 2² = 4
Properties of Limits
- Given real numbers b and c, a positive integer n, and functions f and g with the following limits: lim (x→c) f(x) and lim (x→c) g(x)
- Scalar Multiple: lim (x→c) [b * f(x)] = b * lim (x→c) f(x)
- Sum or Difference: lim (x→c) [f(x) ± g(x)] = lim (x→c) f(x) ± lim (x→c) g(x)
- Product: lim (x→c) [f(x) * g(x)] = lim (x→c) f(x) * lim (x→c) g(x)
- Quotient: lim (x→c) [f(x) / g(x)] = [lim (x→c) f(x)] / [lim (x→c) g(x)], where lim (x→c) g(x) ≠ 0
- Power: lim (x→c) [f(x)]^n = [lim (x→c) f(x)]^n, where n ∈ Z
- Radical: lim (x→c) ⁿ√f(x) = ⁿ√lim (x→c) f(x)
Limit Examples
- Given lim (x→a) f(x) = 4 and lim (x→a) g(x) = -3:
- lim (x→a) 4 * f(x) = 4 * lim (x→a) f(x) = 4 * 4 = 16
- lim (x→a) [3 * f(x) + 5 * g(x)] = 3 * lim (x→a) f(x) + 5 * lim (x→a) g(x) = 3 * 4 + 5 * (-3) = 12 - 15 = -3
- lim (x→a) [f(x) * g(x)] = lim (x→a) f(x) * lim (x→a) g(x) = 4 * (-3) = -12
- lim (x→a) [g(x)]^4 * √f(x) = [lim (x→a) g(x)]^4 * √lim (x→a) f(x) = (-3)^4 * √4 = 81 * 2 = 162
- lim (x→a) [f(x) / g(x)] = [lim (x→a) f(x)] / [lim (x→a) g(x)] = 4 / -3 = -4/3
More Limit Examples
- lim (x→2) (x² + x) = lim (x→2) x² + lim (x→2) x = 2² + 2 = 6
- lim (x→2) [(x + 1)(x - 3)] = lim (x→2) (x + 1) * lim (x→2) (x - 3) = (2 + 1) * (2 - 3) = 3 * (-1) = -3
- lim (x→3) [2(x - 1)] = 2 * lim (x→3) (x - 1) = 2 * (3 - 1) = 4 equivalently, lim (x→3) [2x - 2] = lim (x→3) 2x - lim (x→3) 2 = 2 * 3 - 2 = 4
- lim (x→1) (2x² + x - 3) / (x² + 4) = [2(1)² + 1 - 3] / (1² + 4) = 0/5 = 0
Rationalizing Technique (Conjugate Method)
- Used to evaluate limits involving radicals.
- Example: lim (x→0) (√(x + 1) - 1) / x
- Direct substitution results in an indeterminate form 0/0.
- Multiply by conjugate: lim (x→0) [(√(x + 1) - 1) / x] * [(√(x + 1) + 1) / (√(x + 1) + 1)]
- Simplify: lim (x→0) (x + 1 - 1) / [x(√(x + 1) + 1)] = lim (x→0) x / [x(√(x + 1) + 1)]
- Further simplification: lim (x→0) 1 / (√(x + 1) + 1)
- Evaluate: 1 / (√(0 + 1) + 1) = 1/2
Homework (Rationalizing Technique)
- lim (x→2) (√(4x + 1) - 3) / (x - 2) = 2/3
- lim (x→0) (√(1 + x) - √(1 - x)) / x = 1
- lim (x→-2) (x + 2) / (√(x² + 5) - 3) = 3/2
- lim (x→0) (1 / √(x + 1) - 1 / x) = 1/2
- lim (x→0) (√(x + 5) - √5) / x = √5 / 10
Evaluating Limits by Factoring
- Example: lim (x→0) (5x) / (x² + 2x)
- Factor: lim (x→0) 5x / [x(x + 2)]
- Simplify: lim (x→0) 5 / (x + 2)
- Evaluate: 5 / (0 + 2) = 5/2
Homework (Evaluating Limits by Factoring)
- lim (x→2) (x² + 3x - 10) / (x - 2) = 7
- lim (x→3) (x² - 9) / (x - 3) = 6
- lim (x→2) (x³ - 8) / (x - 2) = 12
- lim (x→4) (4 - x) / (x² - 16) = -1/8
- lim (x→4) (x² - x - 12) / (x² - x - 20) = 7/9
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