Calculus

# Calculus - Chapter 1 Limits and Continuity Chapter 1...

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Chapter 1 Limits and Continuity 29 Chapter 1 Section 1.1 5. a. –2 b. 2 c. does not exist d. 1 e. 1 2 f. –1 g. 3 h. does not exist i. 2 j. 2 7. xf ( x ) ( x ) 1.5 2.22 0.5 1.71 1.1 2.05 0.9 1.95 1.01 2.00 0.99 1.99 1.001 2.00 0.999 2.00 1 lim ( ) 2 x fx + = 1 lim ( ) 2 x = 1 lim ( ) 2 x = 9. 3 0 2 0 on [ 2, 0] by [–3, 0] x 2 1 1 x x + x 2 1 1 x x + 0.9 –1.9 1.1 2.1 0.99 –1.99 1.01 2.01 0.999 1.999 1.001 2.001 2 1 1 lim 2 1 x x x →− =− + 11. 2 0 0 1 on [–2, 0] by [0, 1] x 2 2 2 xx + −− x 2 2 2 + –0.9 0.31 –1.1 0.35 –0.99 0.33 –1.01 0.34 –0.999 0.33 –1.001 0.33 2 2 1 1 lim 3 2 x + = 13. 1 0 1 2 on [–1, 1] by [0, 2] x 2 sin x + x 2 sin x + 0.1 1.10 –0.1 0.90 0.01 1.01 –0.01 0.99 0.001 1.00 –0.001 1.00 2 0 lim 1 sin x x + =

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Chapter 1 Limits and Continuity 30 15. 1 0 1 2 on [–1, 1] by [0, 2] x tan sin x x x tan sin x x 0.1 1.005 –0.1 1.005 0.01 1.000 –0.01 1.000 0.001 1.000 –0.001 1.000 0 tan lim 1 sin x x x = 17. 1 2 3 2 on [1, 3] by [–2, 2] x 2 2 x x x 2 2 x x 2.1 1 1.9 –1 2.01 1 1.99 –1 2.001 1 1.999 –1 2 2 lim 2 x x x does not exist. There is a break in the graph at x = 2. 19. 2 1 0 1 on [–2, 0] by [–1, 1] x 2 2 1 23 x xx +− x 2 2 1 x –0.9 0.05 –1.1 –0.05 –0.99 0.005 –1.01 –0.005 –0.999 0.00 –1.001 0.00 2 2 1 1 lim 0 x x →− = 21. 0 10 2 10 on [0, 2] by [–10, 10] x 2 1 1 x x + x 2 1 1 x x + 1.1 22.1 0.9 –18.1 1.01 202.01 0.99 –198.01 1.001 2002 0.999 –1998 2 1 1 lim 1 x x x + does not exist. There is a break in the graph at x = 1. 23. −1 3 8 0 on [–3, 1] by [–8, 0] x 2 4 2 x x + x 2 4 2 x x + –1.9 3.9 –2.1 –4.1 –1.99 3.99 –2.01 –4.01 –1.999 - 3.999 –2.001 4.001 2 2 4 lim 4 2 x x x =− +
Chapter 1 Limits and Continuity 31 25. 0 10 2 10 on [0, 2] by [–10, 10] x 2 2 1 21 x xx −+ x 2 2 1 x 1.1 21 0.9 –19 1.01 201 0.99 –199 1.001 2001 0.999 –1999 2 2 1 1 lim x x does not exist. There is a break in the graph at x = 1. 27. If g ( a ) = 0 and f ( a ) ±²³´² () lim xa fx gx does not exist. 29. One possibility is x y 1 –1 –3 3 31. One possibility is x y 0 1 3 33. The first argument gives the correct value; it is better than the second argument since it doesn’t depend on specific values of x . 35. x 1 (1 ) x x + x 1 ) x x + 0.1 2.59 –0.1 2.87 0.01 2.70 –0.01 2.73 0.001 2.7169 –0.001 2.7196 1 0 lim(1 ) 2.7182818 x x x +≈ 37. x sec x x 0.1 0.099 0.01 0.010 0.001 0.001 sec 0 lim 0 x x x + = For negative x the values of sec x x are complex numbers. Both the real and the imaginary part of sec x x approach 0 as 0 , x so sec 0 lim 0. x x x = 39. A possible answer is 2 1 if 0 0 x x x x = = −> 41. As x gets arbitrarily close to a , f ( x ) gets arbitrarily close to L . Section 1.2 5. 22 0 lim( 3 1) 0 3(0) 1 1 x −+=− + = 7. 1 lim 2 4 1 2(1) 4 7 x ++= + += 9. 2 52 5 3 lim 8 42 4 x x x −− == ++ 11. 2 33 3 (3 ) (2 ) 6 lim lim lim( 2) 32 5 x x →→ = =+ = 13. 2 2 2 ) ) 2 lim lim ) ) 4 1 lim 2 3 4 x x x x = +− + = + + = + =

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Chapter 1 Limits and Continuity 32 15. 2 3 2 11 2 1 2 (1 ) ( 1 ) 1 lim lim (3 ) ) 23 1 lim 3 1 13 3 4 xx x x x x →→ −+ + = +− ++ = + = + = 17. sin 00 cos 0 sin sin lim lim tan lim cos cos0 1 x x x x x = = = = 19. 2(0) 1 21 22 0 0( ) 0 lim 0 1 10 1 x x e xe x == = 21. 0
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Calculus - Chapter 1 Limits and Continuity Chapter 1...

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