Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
1. Limits and Continuity
Finding Limits Algebraically
Problem 2.7.59
Textbook Question
Use the definitions given in Exercise 57 to prove the following infinite limits.
lim x→1^- 1 / 1 − x=∞
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1
Start by recalling the definition of an infinite limit: lim x→c f(x) = ∞ means that for every M > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then f(x) > M.
In this case, we need to analyze the function f(x) = 1 / (1 - x) as x approaches 1 from the left (x → 1^-).
Observe that as x approaches 1 from the left, the expression 1 - x approaches 0 and is positive, which means that f(x) = 1 / (1 - x) will increase without bound.
To formalize this, choose an arbitrary positive number M and set up the inequality 1 / (1 - x) > M, which simplifies to 1 > M(1 - x).
Solve for x to find the corresponding δ: rearranging gives x > 1 - 1/M, and thus, we can choose δ = 1 - 1/M to ensure that |x - 1| < δ leads to f(x) > M.
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