Solution Manual For Calculus & Its Applications, 3rd Edition by Marvin L. Bittinger David J. Ellenbo

Page 1

Chapter 1 Differentiation 13. The notation lim is read “the limit, as x x 2

Exercise Set 1.1 1.

2.

3.

4.

5.

approaches 2 from the right.”

The limit of the sequence is 2.8 and using the notation will give the expression x  2.8.

x 3

approaches 3 from the left”.

The limit of the sequence is 0.3 and using the notation will give the expression x  0.3.

15. The notation lim is read “the limit, as x

The limit of the sequence is 3 and using the notation will give the expression x  3.

16. The notation lim1 is read “the limit, as x

The limit of the sequence is 4.9 and using the notation will give the expression x  4.9 . The limit of the sequence is

2 and using the 3

notation will give the expression x 

6.

14. The notation lim is read “the limit, as x

The limit of the sequence is

2 . 3

4 and using the 3

notation will give the expression x 

4 . 3

7.

The limit of the sequence is 5.4 and using the notation will give the expression x  5.4.

8.

The limit of the sequence is 0.3 and using the notation will give the expression x  0.3.

9.

The limit of the sequence is 1 and using the notation will give the expression x  1.

10. The limit of the sequence is 0 and using the notation will give the expression x  0. 11. As x approaches 0, the value of x  5 approaches 5. 12. We solve the equation: 2 x  6 x  3 Therefore, As x approaches 3, the value of x approaches 6.

x 5

approaches 5”. x 2

approaches

1 ”. 2

17. The notation lim f  x  is read “the limit, as x x4

approaches 4, of f  x  . ” 18. The notation lim g  x  is read “the limit, as x x 1

approaches 1, of g  x  . ” 19. The notation lim F  x  is read “the limit, as x x 5

approaches 5 from the left, of F  x  . ” 20. The notation lim G  x  is read “the limit, as x x4

approaches 4 from the right, of G  x  . ” 21. a) As inputs x approach 1 from the left, outputs f  x  approach 3. Thus the limit from the left is 3. That is, lim  f  x   3. x 1

b) As inputs x approach 1 from the right, outputs f  x  approach 3. That is, lim f  x   3.

x 1

c) From parts (a) and (b) we find lim f  x   3. x 1

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