This preview shows pages 1–3. Sign up to view the full content.

Chapter 2 Review Exercises (pp. 91–93) 1. lim x 2 2 ( x 3 2 2 x 2 1 1) 5 ( 2 2) 3 2 2( 2 2) 2 1 1 52 15 2. lim x →-2 } 3 x 2 x 2 2 1 2 x 1 1 5 } 55 } 2 5 1 } 3. No limit, because the expression ˇ 1 w 2 w 2 w x w is undefined for values of x near 4. 4. No limit, because the expression ˇ 4 9 w 2 w x w 2 w is undefined for values of x near 5. 5. lim x 0 5 lim x 0 } 2 2 2 x (2 (2 1 1 x ) x ) }5 lim x 0 } 2 x (2 2 1 x x ) } 5 lim x 0 1 2 } 2(2 1 1 x ) } 2 52 } 2(2 1 1 0) }52} 1 4 } 6. lim x 6‘ } 2 5 x x 2 2 1 1 3 7 } 5 lim x 6‘ } 2 5 x x 2 2 } 5 } 2 5 } 7. An end behavior model for } 12 x x 4 3 1 1 x 1 3 28 } is } 1 x 2 4 x 3 } 5 } 1 1 2 } x . Therefore: lim x } 12 x x 4 3 1 1 x 1 3 28 lim x } 1 1 2 } x 5‘ lim x 2‘ } 12 x x 4 3 1 1 x 1 3 28 lim x 2‘ } 1 1 2 } x 52‘ 8. lim x 0 } sin 4 x 2 x } 5 } 1 2 } lim x 0 } sin 2 x 2 x } 5 } 1 2 } (1) 5 } 1 2 } 9. Multiply the numerator and denominator by sin x . lim x 0 } x c x sc cs x c 1 x 1 lim x 0 } x 1 x sin x } 5 lim x 0 1 1 1 } sin x x } 2 5 1 lim x 0 1 2 1 1 lim x 0 } sin x x } 2 5 1 1 1 5 2 10. lim x 0 e x sin x 5 e 0 sin 0 5 1 ? 0 5 0 11. Let x 5 } 7 2 } 1 h , where h is in 1 0, } 1 2 } 2 . Then int (2 x 2 1) 5 int 3 2 1 } 7 2 } 2 1 2 h – 1 4 5 int (6 1 2 h ) 5 6, because 6 1 2 h is in (6, 7). Therefore, lim x 7/2 1 int (2 x – 1) 5 lim x 7/2 1 6 5 6. 12. Let x 5 } 7 2 } 1 h , where h is in 1 2} 1 2 } ,0 2 . Then int (2 x 2 1) 5 int 3 2 1 } 7 2 } 2 1 2 h 2 1 4 5 int (6 1 2 h ) 5 5, because 6 1 2 h is in (5, 6). Therefore, lim x 7/2 2 int (2 x 2 1) 5 lim x 7/2 2 5 5 5 13. Since lim x ( 2 e 2 x ) 5 lim x e 2 x 5 0, and 2 e 2 x # e 2 x cos x # e 2 x for all x , the Sandwich Theorem gives lim x e 2 x cos x 5 0. 14. Since the expression x is an end behavior model for both x 1 sin x and x 1 cos x ,lim x 5 lim x } x x } 5 1. 15. Limit exists. 16. Limit exists. 17. Limit exists. 18. Limit does not exist. 19. Limit exists. 20. Limit exists. 21. Yes 22. No 23. No 24. 25. (a) lim x 3 2 g ( x ) 5 1 ( b) g (3) 5 1.5 ( c) No, since lim x 3 2 g ( x ) ± g (3). ( d) g is discontinuous at x 5 3 (and at points not in the domain). ( e) Yes, the discontinuity at x 5 3 can be removed by assigning the value 1 to g (3). 26. (a) lim x 1 2 k ( x ) 5 1.5 ( b) lim x 1 1 k ( x ) 5 0 ( c) k (1) 5 0 ( d) No, since lim x 1 2 k ( x ) ± k (1) ( e) k is discontinuous at x 5 1 (and at points not in the domain). ( f) No, the discontinuity at x 5 1 is not removable because the one-sided limits are different. 27. [ 2 4, 4] by [ 2 3, 3] (a) Vertical asymptote: x 2 ( b) Left-hand limit 5 lim x –2 2 } x x 1 1 3 2 }52‘ Right-hand limit: lim x –2 1 } x x 1 1 3 2 }5‘ x 1 sin x } x 1 cos x } 2 1 1 x } 2 } 1 2 } }} x ( 2 2) 2 1 1 }}} 3( 2 2) 2 2 2( 2 2) 1 5 66 Chapter 2 Review

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
28. [ 2 4, 4] by [ 2 3, 3] (a) Vertical asymptotes: x 5 0, x 52 2 ( b) At x 5 0: Left-hand limit 5 lim x 0 2 } x 2 x ( x 2 1 1 2) }52‘ Right-hand limit 5 lim x 0 1 } x 2 x ( x 2 1 1 2) At x 2: Left-hand limit 5 lim x –2 2 } x 2 x ( x 2 1 1 2) }5‘ Right-hand limit 5 lim x –2 1 } x 2 x ( x 2 1 1 2) 29. (a) At x 52 1: Left-hand limit 5 lim x 2 1 2 f ( x ) 5 lim x 2 1 2 (1) 5 1 Right-hand limit 5 lim x 2 1 1 f ( x ) 5 lim x 2 1 1 ( 2 x ) 5 1 At x 5 0: Left-hand limit 5 lim x 0 2 f ( x ) 5 lim x 0 2 ( 2 x ) 5 0 Right-hand limit 5 lim x 0
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}