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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).

lim x 3 x 2 3 x x 2 9 ,

x= –2.5, –2.9, –2.95, –2.99, –2.999, –2.9999, –3.5, –3.1, –3.05, –3.01, –3.001, –3.0001

To determine

To guess: The value of limx3x23xx29.

Explanation

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

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