Precalculus with Limits: A Graphing Approach
Precalculus with Limits: A Graphing Approach
7th Edition
ISBN: 9781305071711
Author: Ron Larson
Publisher: Brooks/Cole
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Chapter 11.2, Problem 40E
To determine

To find: The value of the limx2ln(x1)x2 using table feature of a graphing utility. Also confirm your result graphically.

Expert Solution & Answer
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Answer to Problem 40E

The value of the limits is limx2ln(x1)x2=1

Explanation of Solution

Given information:

Given expression of limit as limx2ln(x1)x2

Let, f(x)=ln(x1)x2

The value of the function at x=2.1 is,

  f(2.1)=ln(2.11)2.12=ln(1.1)0.1=0.9531

The value of the function at x=2.01 is,

  f(2.01)=ln(2.011)2.012=ln(1.01)0.01=0.9950

The value of the function at x=2.001 is,

  f(2.001)=ln(2.0011)2.0012=ln(1.001)0.001=0.99950

The value of the function at x=2 is,

  f(2)=ln(21)22=00=undefined

The value of the function at x=1.9 is,

  f(1.9)=ln(1.91)1.92=ln(0.9)0.1=1.054

The value of the function at x=1.99 is,

  f(1.99)=ln(1.991)1.992=ln(0.99)0.01=1.00503

The value of the function at x=1.999 is,

  f(1.999)=ln(1.9991)1.9992=ln(0.999)0.001=1.00050

Hence, the complete table is,

  x2.12.012.00121.91.991.999f(x)0.95310.99500.99950undefined1.0541.005031.0005

The value of the limit is, limx2ln(x1)x2=1

Graph of the function can be drawn as,

  Precalculus with Limits: A Graphing Approach, Chapter 11.2, Problem 40E

From the graph it is clear that limx2ln(x1)x2=1

Chapter 11 Solutions

Precalculus with Limits: A Graphing Approach

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