Explanation Given Information: The provided limit notation is lim x → 0 f ( x ) , where, f ( x ) = { x + 2 if x < 0 3 x + 2 if x ≥ 0 } . Calculation: Consider the provided limit notation, lim x → 0 f ( x ) , where, f ( x ) = { x + 2 if x < 0 3 x + 2 if x ≥ 0 } In making the table, choose the value of x close to 0 from the left and from the right as x approaches 0. As x approaches 0 from the left arbitrarily start with x = − 0.01 . Then select two additional values of x that are closer to 0, but still less than 0, we choose − 0.001 and − 0.0001 . For x < 0 , use the function f ( x ) = x + 2 . Thus, the limit notation becomes lim x → 0 x + 2 . Evaluate f at each chosen value of x to obtain the corresponding values of f ( x ) . All the values of x are in radian. At x = − 0.01 , Substitute, x = − 0.01 in the provided limit notation, lim x → 0 x + 2 . Therefore, lim x → − 0.01 x + 2 = − 0.01 + 2 = 1.99 Now, at x = − 0.001 Substitute, x = − 0.001 in the provided limit notation, lim x → 0 x + 2 . lim x → − 0.001 x + 2 = − 0.001 + 2 = 1.999 At x = − 0.0001 Substitute, x = − 0.0001 in the provided limit notation, lim x → 0 x + 2 . lim x → − 0.0001 x + 2 = − 0.0001 + 2 = 1.9999 Now, as x approaches 0 from the right arbitrarily start with x = 0.01 . Then select two additional values of x that are closer to 0 , but still greater than 0 , we choose 0.001 and 0.0001 . For x ≥ 0 , use the function f ( x ) = 3 x + 2

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ISBN: 9780135963173

Author: Blitzer

Publisher: PEARSON CO

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When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topic…

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Chapter 11.1, Problem 18PE

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To calculate: The value of limit, limx→0f(x), where, f(x)={x+2 if x<03x+2 if x≥0} by constructing a table for different values of x.

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Chapter 11.1, Problem 17PE

Chapter 11.1, Problem 19PE

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Chapter 11 Solutions

EP PRECALCULUS-MYLABMATH+ETEXT ACCESS

Ch. 11.1 - Check Point 1 Find: . Ch. 11.1 - Prob. 2CP Ch. 11.1 - Prob. 3CP Ch. 11.1 - Prob. 4CP Ch. 11.1 - Prob. 5CP Ch. 11.1 - Prob. 1CVC Ch. 11.1 - Prob. 2CVC Ch. 11.1 - Prob. 3CVC Ch. 11.1 - Fill in each blank so that the resulting statement...Ch. 11.1 - Fill in each blank so that the resulting statement...

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The value of limit, lim x → 0 f ( x ) , where, f ( x ) = { x + 2 if x < 0 3 x + 2 if x ≥ 0 } by constructing a table for different values of x .

Solution Summary: The author calculates the value of the provided limit notation by constructing a table for different values of x.

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To calculate: The value of limit, limx→0f(x), where, f(x)={x+2 if x<03x+2 if x≥0} by constructing a table for different values of x.

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Chapter 11 Solutions

The value of limit, lim x → 0 f ( x ) , where, f ( x ) = { x + 2 if x &lt; 0 3 x + 2 if x ≥ 0 } by constructing a table for different values of x .

Solution Summary: The author calculates the value of the provided limit notation by constructing a table for different values of x.

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To calculate: The value of limit, limx→0f(x), where, f(x)={x+2 if x<03x+2 if x≥0} by constructing a table for different values of x.

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Chapter 11 Solutions

The value of limit, lim x → 0 f ( x ) , where, f ( x ) = { x + 2 if x < 0 3 x + 2 if x ≥ 0 } by constructing a table for different values of x .