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 3.5.82
Textbook Question
Another method for proving lim x→0 cos x−1/x = 0 Use the half-angle formula sin²x = 1− cos 2x/2 to prove that lim x→0 cos x−1/x=0.
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Start by rewriting the limit expression lim x→0 (cos x - 1)/x to analyze its behavior as x approaches 0.
Recognize that both the numerator (cos x - 1) and the denominator (x) approach 0 as x approaches 0, indicating that we have a 0/0 indeterminate form.
Use the half-angle formula sin²x = (1 - cos 2x)/2 to express cos x - 1 in terms of sin²x, noting that cos x = 1 - 2sin²(x/2).
Substitute this expression into the limit, transforming it into lim x→0 (-2sin²(x/2))/x, which can be simplified further.
Apply the limit properties and L'Hôpital's Rule if necessary, or use the small angle approximation sin(x) ≈ x to evaluate the limit as x approaches 0.
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