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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Hass 15th EditionCh. 2 - Limits and Continuity문제 2.P.1
2장, 문제 2.P.1

Limits and Continuity


Graph the function


1 , x ≤ ―1
―x , ―1 < x < 0
ƒ(x) = { 1 , x = 0 ,
―x , 0 < x < 1
1 , x ≥ 1


Then discuss, in detail, limits, one-sided limits, continuity, and one-sided continuity of ƒ at x = ―1 , 0 , and 1. Are any of the discontinuities removable? Explain.

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1
Step 1: Identify the piecewise function and its components. The function ƒ(x) is defined as follows: ƒ(x) = 1 for x ≤ -1, ƒ(x) = -1/x for -1 < x < 0, ƒ(x) = 1 for x = 0, ƒ(x) = -x for 0 < x < 1, and ƒ(x) = 1 for x ≥ 1.
Step 2: Analyze the limits and one-sided limits at x = -1. Evaluate the left-hand limit as x approaches -1 from the left (x → -1⁻) and the right-hand limit as x approaches -1 from the right (x → -1⁺). Compare these limits to the value of the function at x = -1 to determine continuity.
Step 3: Analyze the limits and one-sided limits at x = 0. Evaluate the left-hand limit as x approaches 0 from the left (x → 0⁻) and the right-hand limit as x approaches 0 from the right (x → 0⁺). Compare these limits to the value of the function at x = 0 to determine continuity.
Step 4: Analyze the limits and one-sided limits at x = 1. Evaluate the left-hand limit as x approaches 1 from the left (x → 1⁻) and the right-hand limit as x approaches 1 from the right (x → 1⁺). Compare these limits to the value of the function at x = 1 to determine continuity.
Step 5: Discuss the types of discontinuities at x = -1, 0, and 1. Determine if any discontinuities are removable by checking if redefining the function at these points can make the function continuous. A discontinuity is removable if the limit exists at that point, but the function value is different or undefined.

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이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Limits

Limits describe the behavior of a function as the input approaches a certain value. For the function ƒ(x), understanding limits at x = -1, 0, and 1 involves evaluating the function's value as x approaches these points from both sides. This helps determine if the function approaches a specific value or diverges, which is crucial for analyzing continuity and discontinuities.
추천 영상:
05:50
One-Sided Limits

One-Sided Limits

One-sided limits consider the behavior of a function as the input approaches a specific value from one side only, either from the left (x → c⁻) or the right (x → c⁺). For ƒ(x), examining one-sided limits at x = -1, 0, and 1 helps identify potential discontinuities and whether the function behaves differently when approaching these points from different directions.
추천 영상:
05:50
One-Sided Limits

Continuity and Discontinuity

A function is continuous at a point if the limit exists and equals the function's value at that point. Discontinuities occur when this condition fails. For ƒ(x), analyzing continuity at x = -1, 0, and 1 involves checking if the limits and function values match. Removable discontinuities can be fixed by redefining the function at the point, while non-removable ones indicate a more fundamental break in the function's behavior.
추천 영상:
05:34
Intro to Continuity
관련 실천
교과서 질문

Limits and Infinity


Find the limits in Exercises 37–46.


2x² + 3

lim -------------

x→⁻∞ 5x² + 7

273
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교과서 질문

Limits of Rational Functions


In Exercises 13–22, find the limit of each rational function (a) as x → ∞ and (b) as x → −∞. Write ∞ or −∞ where appropriate.


f(x) = (x + 1)/(x² + 3)

365
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교과서 질문

Limits and Continuity

Repeat the instructions of Exercise 1 for


1 , x ≤ ―1

1/x , 0 < |x| < 1

ƒ(x) = { 0, x = 1 ,

1 , x > 1 .

241
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교과서 질문

Finding Limits of Differences When x → ±∞


Find the limits in Exercises 84–90. (Hint: Try multiplying and dividing by the conjugate.)


lim x → ∞ (√(9x² − x) − 3x)

270
views
교과서 질문

Limits and Continuity

On what intervals are the following functions continuous?


a. ƒ(x) = tan x

266
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교과서 질문

Suppose that limx→−2 p(x) = 4, limx→−2 r(x) = 0, and limx→−2 s(x) = −3. Find


a. limx→−2 (p(x) + r(x) + s(x))

193
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