- 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 3h
Textbook Question
Limits and Continuity
Suppose that ƒ(t) and ƒ(t) are defined for all t and that lim t → t₀ ƒ(t) = ―7 and lim (t → t₀) g (t) = 0 . Find the limit as t → t₀ of the following functions.
h. 1 / ƒ(t)

1
To find the limit of 1/ƒ(t) as t approaches t₀, we start by considering the given limit of ƒ(t) as t approaches t₀, which is -7.
Since lim t → t₀ ƒ(t) = -7, this means that as t gets closer to t₀, the value of ƒ(t) gets closer to -7.
The function 1/ƒ(t) is continuous wherever ƒ(t) is not zero. Since -7 is not zero, 1/ƒ(t) is continuous at t₀.
By the property of limits, if a function is continuous at a point and the limit of the function's argument exists at that point, then the limit of the function is simply the function evaluated at the limit of the argument.
Therefore, the limit of 1/ƒ(t) as t approaches t₀ is 1 divided by the limit of ƒ(t) as t approaches t₀, which is 1/(-7).
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