(a) Answer The value of limit of f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 as x → − ∞ is 1 . Explanation Given information: The function is f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 . Calculation: Consider the given function f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 . For x → − ∞ , the function is f ( x ) = x − 2 x − 1 . So, lim x → − ∞ f ( x ) = lim x → − ∞ ( x − 2 x − 1 ) = lim x → − ∞ ( 1 − 1 x − 1 ) = 1 Therefore, the value of limit of f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 as x → − ∞ is 1 . (b) To determine To find: The value of limit of f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 as x → ∞ . Answer The value of limit of f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 as x → ∞ is 0 . Explanation Given information: The given function is f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 . Calculation: Consider the given function f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 . For x → ∞ , the function is f ( x ) = 1 x 2 . So, lim x → ∞ f ( x ) = lim x → ∞ ( 1 x 2 ) = 0 Therefore, the value of limit of f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 as x → ∞ is 0 . (c) To determine To find: The value of limit of f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 as x → 0 − . Answer The value of limit of f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 as x → 0 − is 2 . Explanation Given information: The given function is f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 . Calculation: As x approaches to 0 from the left side, the function is f ( x ) = x − 2 x − 1 . The value of limit is: lim x → 0 − f ( x ) = lim x → 0 − ( x − 2 x − 1 ) = 0 − 2 0 − 1 = 2 Therefore, the value of limit of f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 as x → 0 − is 2 . (d) To determine To find: The value of limit of f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 as x → 0 + . Answer The value of limit of f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 as x → 0 + is ∞ . Explanation Given information: The given function is f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 . Calculation: As x approaches to 0 from the right side, the function is f ( x ) = 1 x 2 . The value of limit is: lim x → 0 − f ( x ) = lim x → 0 − ( 1 x 2 ) = 1 0 = ∞ Therefore, the value of limit of f ( x ) = { x − 2 x − 1 , x ≤ 0 1 x 2 , x > 0 as x → 0 + is ∞ .

Calculus: Graphical, Numerical, Algebraic: Solutions Manual

3rd Edition

ISBN: 9780132014144

Author: Ross L. Finney, Franklin Demana, Bert K. Waits, Daniel Kennedy

Publisher: Pearson Prentice Hall

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Limits

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Question

Chapter 2.2, Problem 54E

(a)

To determine

To find: The value of limit of f(x)={x−2x−1, x≤01x2, x>0 as x→−∞ .

(a)

Expert Solution

Answer to Problem 54E

The value of limit of f(x)={x−2x−1, x≤01x2, x>0 as x→−∞ is 1 .

Explanation of Solution

Given information:

The function is f(x)={x−2x−1, x≤01x2, x>0 .

Calculation:

Consider the given function f(x)={x−2x−1, x≤01x2, x>0 .

For x→−∞ , the function is f(x)=x−2x−1 . So,

limx→−∞f(x)=limx→−∞(x−2x−1)=limx→−∞(1−1x−1)=1

Therefore, the value of limit of f(x)={x−2x−1, x≤01x2, x>0 as x→−∞ is 1 .

(b)

To determine

To find:The value of limit of f(x)={x−2x−1, x≤01x2, x>0 as x→∞ .

(b)

Expert Solution

Answer to Problem 54E

The value of limit of f(x)={x−2x−1, x≤01x2, x>0 as x→∞ is 0 .

Explanation of Solution

Given information:

The given function is f(x)={x−2x−1, x≤01x2, x>0 .

Calculation:

Consider the given function f(x)={x−2x−1, x≤01x2, x>0 .

For x→∞ , the function is f(x)=1x2 . So,

limx→∞f(x)=limx→∞(1x2)=0

Therefore, the value of limit of f(x)={x−2x−1, x≤01x2, x>0 as x→∞ is 0 .

(c)

To determine

To find:The value of limit of f(x)={x−2x−1, x≤01x2, x>0 as x→0− .

(c)

Expert Solution

Answer to Problem 54E

The value of limit of f(x)={x−2x−1, x≤01x2, x>0 as x→0− is 2 .

Explanation of Solution

Given information:

The given function is f(x)={x−2x−1, x≤01x2, x>0 .

Calculation:

As x approaches to 0 from the left side, the function is f(x)=x−2x−1 .

The value of limit is:

limx→0−f(x)=limx→0−(x−2x−1)=0−20−1=2

Therefore, the value of limit of f(x)={x−2x−1, x≤01x2, x>0 as x→0− is 2 .

(d)

To determine

To find:The value of limit of f(x)={x−2x−1, x≤01x2, x>0 as x→0+ .

(d)

Expert Solution

Answer to Problem 54E

The value of limit of f(x)={x−2x−1, x≤01x2, x>0 as x→0+ is ∞ .

Explanation of Solution

Given information:

The given function is f(x)={x−2x−1, x≤01x2, x>0 .

Calculation:

As x approaches to 0 from the right side, the function is f(x)=1x2 .

The value of limit is:

limx→0−f(x)=limx→0−(1x2)=10=∞

Therefore, the value of limit of f(x)={x−2x−1, x≤01x2, x>0 as x→0+ is ∞ .

Chapter 2.2, Problem 53E

Chapter 2.2, Problem 55E

Chapter 2 Solutions

Calculus: Graphical, Numerical, Algebraic: Solutions Manual

Ch. 2.1 - Prob. 1QR Ch. 2.1 - Prob. 2QR Ch. 2.1 - Prob. 3QR Ch. 2.1 - Prob. 4QR Ch. 2.1 - Prob. 5QR Ch. 2.1 - Prob. 6QR Ch. 2.1 - Prob. 7QR Ch. 2.1 - Prob. 8QR Ch. 2.1 - Prob. 9QR Ch. 2.1 - Prob. 10QR

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.1 - Prob. 41E Ch. 2.1 - Prob. 42E Ch. 2.1 - Prob. 43E Ch. 2.1 - Prob. 44E Ch. 2.1 - Prob. 45E Ch. 2.1 - Prob. 46E Ch. 2.1 - Prob. 47E Ch. 2.1 - Prob. 48E Ch. 2.1 - Prob. 49E Ch. 2.1 - Prob. 50E Ch. 2.1 - 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Prob. 62E Ch. 2.2 - Prob. 63E Ch. 2.2 - Prob. 64E Ch. 2.2 - Prob. 65E Ch. 2.2 - Prob. 66E Ch. 2.2 - Prob. 67E Ch. 2.2 - Prob. 68E Ch. 2.2 - Prob. 69E Ch. 2.2 - Prob. 70E Ch. 2.2 - Prob. 71E Ch. 2.2 - Prob. 1QQ Ch. 2.2 - Prob. 2QQ Ch. 2.2 - Prob. 3QQ Ch. 2.2 - Prob. 4QQ Ch. 2.3 - Prob. 1QR Ch. 2.3 - Prob. 2QR Ch. 2.3 - Prob. 3QR Ch. 2.3 - Prob. 4QR Ch. 2.3 - Prob. 5QR Ch. 2.3 - Prob. 6QR Ch. 2.3 - Prob. 7QR Ch. 2.3 - Prob. 8QR Ch. 2.3 - Prob. 9QR Ch. 2.3 - Prob. 10QR Ch. 2.3 - Prob. 1E Ch. 2.3 - Prob. 2E Ch. 2.3 - Prob. 3E Ch. 2.3 - Prob. 4E Ch. 2.3 - Prob. 5E Ch. 2.3 - Prob. 6E Ch. 2.3 - Prob. 7E Ch. 2.3 - Prob. 8E Ch. 2.3 - Prob. 9E Ch. 2.3 - Prob. 10E Ch. 2.3 - Prob. 11E Ch. 2.3 - Prob. 12E Ch. 2.3 - Prob. 13E Ch. 2.3 - Prob. 14E Ch. 2.3 - Prob. 15E Ch. 2.3 - Prob. 16E Ch. 2.3 - Prob. 17E Ch. 2.3 - Prob. 18E Ch. 2.3 - Prob. 19E Ch. 2.3 - Prob. 20E Ch. 2.3 - Prob. 21E Ch. 2.3 - Prob. 22E Ch. 2.3 - Prob. 23E Ch. 2.3 - Prob. 24E Ch. 2.3 - Prob. 25E Ch. 2.3 - Prob. 26E Ch. 2.3 - Prob. 27E Ch. 2.3 - Prob. 28E Ch. 2.3 - Prob. 29E Ch. 2.3 - Prob. 30E Ch. 2.3 - Prob. 31E Ch. 2.3 - Prob. 32E Ch. 2.3 - Prob. 33E Ch. 2.3 - Prob. 34E Ch. 2.3 - Prob. 35E Ch. 2.3 - Prob. 36E Ch. 2.3 - Prob. 37E Ch. 2.3 - Prob. 38E Ch. 2.3 - Prob. 39E Ch. 2.3 - Prob. 40E Ch. 2.3 - Prob. 41E Ch. 2.3 - Prob. 42E Ch. 2.3 - Prob. 43E Ch. 2.3 - Prob. 44E Ch. 2.3 - Prob. 45E Ch. 2.3 - Prob. 46E Ch. 2.3 - Prob. 47E Ch. 2.3 - Prob. 48E Ch. 2.3 - Prob. 49E Ch. 2.3 - Prob. 50E Ch. 2.3 - Prob. 51E Ch. 2.3 - Prob. 52E Ch. 2.3 - Prob. 53E Ch. 2.3 - Prob. 54E Ch. 2.3 - Prob. 55E Ch. 2.3 - Prob. 56E Ch. 2.3 - Prob. 57E Ch. 2.3 - Prob. 58E Ch. 2.3 - Prob. 59E Ch. 2.3 - Prob. 60E Ch. 2.3 - Prob. 61E Ch. 2.3 - Prob. 62E Ch. 2.3 - Prob. 63E Ch. 2.3 - Prob. 64E Ch. 2.4 - Prob. 1QR Ch. 2.4 - Prob. 2QR Ch. 2.4 - Prob. 3QR Ch. 2.4 - Prob. 4QR Ch. 2.4 - Prob. 5QR Ch. 2.4 - Prob. 6QR Ch. 2.4 - Prob. 7QR Ch. 2.4 - Prob. 8QR Ch. 2.4 - Prob. 9QR Ch. 2.4 - Prob. 10QR Ch. 2.4 - Prob. 1E Ch. 2.4 - Prob. 2E Ch. 2.4 - Prob. 3E Ch. 2.4 - Prob. 4E Ch. 2.4 - Prob. 5E Ch. 2.4 - Prob. 6E Ch. 2.4 - Prob. 7E Ch. 2.4 - Prob. 8E Ch. 2.4 - Prob. 9E Ch. 2.4 - Prob. 10E Ch. 2.4 - Prob. 11E Ch. 2.4 - Prob. 12E Ch. 2.4 - Prob. 13E Ch. 2.4 - Prob. 14E Ch. 2.4 - Prob. 15E Ch. 2.4 - Prob. 16E Ch. 2.4 - Prob. 17E Ch. 2.4 - Prob. 18E Ch. 2.4 - Prob. 19E Ch. 2.4 - Prob. 20E Ch. 2.4 - Prob. 21E Ch. 2.4 - Prob. 22E Ch. 2.4 - Prob. 23E Ch. 2.4 - Prob. 24E Ch. 2.4 - Prob. 25E Ch. 2.4 - Prob. 26E Ch. 2.4 - Prob. 27E Ch. 2.4 - Prob. 28E Ch. 2.4 - Prob. 29E Ch. 2.4 - Prob. 30E Ch. 2.4 - Prob. 31E Ch. 2.4 - Prob. 32E Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.4 - Prob. 37E Ch. 2.4 - Prob. 38E Ch. 2.4 - Prob. 39E Ch. 2.4 - Prob. 40E Ch. 2.4 - Prob. 41E Ch. 2.4 - Prob. 42E Ch. 2.4 - Prob. 43E Ch. 2.4 - Prob. 44E Ch. 2.4 - Prob. 45E Ch. 2.4 - Prob. 46E Ch. 2.4 - Prob. 47E Ch. 2.4 - Prob. 48E Ch. 2.4 - Prob. 49E Ch. 2.4 - Prob. 50E Ch. 2.4 - Prob. 51E Ch. 2.4 - Prob. 52E Ch. 2.4 - Prob. 53E Ch. 2.4 - Prob. 54E Ch. 2.4 - Prob. 55E Ch. 2.4 - Prob. 1QQ Ch. 2.4 - Prob. 2QQ Ch. 2.4 - Prob. 3QQ Ch. 2.4 - Prob. 4QQ Ch. 2 - Prob. 1RE Ch. 2 - Prob. 2RE Ch. 2 - Prob. 3RE Ch. 2 - Prob. 4RE Ch. 2 - Prob. 5RE Ch. 2 - Prob. 6RE Ch. 2 - Prob. 7RE Ch. 2 - Prob. 8RE Ch. 2 - Prob. 9RE Ch. 2 - Prob. 10RE Ch. 2 - Prob. 11RE Ch. 2 - Prob. 12RE Ch. 2 - Prob. 13RE Ch. 2 - Prob. 14RE Ch. 2 - Prob. 15RE Ch. 2 - Prob. 16RE Ch. 2 - Prob. 17RE Ch. 2 - Prob. 18RE Ch. 2 - Prob. 19RE Ch. 2 - Prob. 20RE Ch. 2 - Prob. 21RE Ch. 2 - Prob. 22RE Ch. 2 - Prob. 23RE Ch. 2 - Prob. 24RE Ch. 2 - Prob. 25RE Ch. 2 - Prob. 26RE Ch. 2 - Prob. 27RE Ch. 2 - Prob. 28RE Ch. 2 - Prob. 29RE Ch. 2 - Prob. 30RE Ch. 2 - Prob. 31RE Ch. 2 - Prob. 32RE Ch. 2 - Prob. 33RE Ch. 2 - Prob. 34RE Ch. 2 - Prob. 35RE Ch. 2 - Prob. 36RE Ch. 2 - Prob. 37RE Ch. 2 - Prob. 38RE Ch. 2 - Prob. 39RE Ch. 2 - Prob. 40RE Ch. 2 - Prob. 41RE Ch. 2 - Prob. 42RE Ch. 2 - Prob. 43RE Ch. 2 - Prob. 44RE Ch. 2 - Prob. 45RE Ch. 2 - Prob. 46RE Ch. 2 - Prob. 47RE Ch. 2 - Prob. 48RE Ch. 2 - Prob. 49RE Ch. 2 - 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