Textbook Question
Horizontal and Vertical Asymptotes
Use limits to determine the equations for all horizontal asymptotes.
_________
/ x² + 9
d. y = / -------------
√ 9x² + 1
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1. Limits and Continuity
Finding Limits Algebraically
2:04 minutesProblem 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)
Verified step by step guidance1
Identify the given limits: lim t → t₀ ƒ(t) = -7 and lim t → t₀ g(t) = 0. We need to find lim t → t₀ of 1/ƒ(t).
Recall the limit property for the reciprocal function: If lim t → t₀ ƒ(t) = L and L ≠ 0, then lim t → t₀ 1/ƒ(t) = 1/L.
Apply the property to the given function: Since lim t → t₀ ƒ(t) = -7 and -7 ≠ 0, we can use the reciprocal limit property.
Substitute the limit value into the reciprocal property: lim t → t₀ 1/ƒ(t) = 1/(-7).
Conclude that the limit of 1/ƒ(t) as t approaches t₀ is the reciprocal of -7, which is -1/7.
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Video duration:
2mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits
A limit describes the value that a function approaches as the input approaches a certain point. In this context, the limit of ƒ(t) as t approaches t₀ is given as -7, indicating that as t gets closer to t₀, ƒ(t) gets closer to -7. Understanding limits is crucial for analyzing the behavior of functions near specific points, especially when dealing with continuity and discontinuity.
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Continuity
A function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. In this case, since lim t → t₀ ƒ(t) = -7, we can infer that ƒ(t) is continuous at t₀ if ƒ(t₀) is also -7. Continuity is essential for ensuring that limits can be evaluated without encountering undefined behavior.
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Reciprocal Limits
The limit of the reciprocal of a function, such as 1/ƒ(t), can be evaluated using the limit of the function itself. If lim t → t₀ ƒ(t) = -7, then lim t → t₀ (1/ƒ(t)) = 1/(-7) = -1/7, provided that ƒ(t) does not approach zero. This concept is important for understanding how limits behave under operations like taking reciprocals.
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