BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 2.2, Problem 18E
To determine

To guess: The value of limx3x23xx29.

Expert Solution

Answer to Problem 18E

The limit of the given function does not exist.

Explanation of Solution

Given:

The values of x are 2.5,2.9,2.95,2.99,2.999,2.9999,3.5,3.1,3.05,3.01,

3.001 and 3.0001.

Calculation:

Since x approaches −3, the values 3.5,3.1,3.05,3.01,3.001 and 3.0001 are to the left of −3 and the values 2.5,2.9,2.95,2.99,2.999, and2.9999 are to the right of −3.

Evaluate the function (correct to 6 decimal places) when x=3.5,3.1,3.05,3.01,3.001 and 3.0001 as shown in the below table.

xx23xx29f(x)=x23xx29
−3.522.753.257
−3.118.910.6131
−3.0518.45250.302561
−3.0118.09010.0601301
−3.00118.0090.0060013,001
−3.000118.00090.000630,001

Here, the value of f(x) approaches the large negative value as x closer to −3 from the left side.

That is, limx3x23xx29=.

Evaluate the function (correct to 6 decimal places) when x=2.5,2.9,2.95,2.99,2.999, and2.9999 as shown in the below table.

xx23xx29f(x)=x23xx29
−2.513.75−2.75−5
−2.917.11−0.59−29
−2.9517.5525−0.2975−59
−2.9917.9101−0.0599−299
−2.99917.991−0.006−2,999
−2.999917.9991−0.0006−29,999

Here, the value of f(x) gets closer to the largest positive value as x approaches to 0 from the right side.

That is, limx3+x23xx29=.

Since the left-hand limit and the right-hand limit are not the same, the limit of the function does not exist.

Thus, it can be guessed that limx3x23xx29 does not exist.

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