Unformatted text preview: Section 8.1
5. 3. 323 [0, 2] by [0, 3] [0, 3] by [0, 3] As t1, 6. t t 1 1 approaches 2. From the graph, the limit appears to be 1. The limit leads to the indeterminate form 0.
1 x x
1 x ln 1 ln 1 x ln 1 1 x
1 1 1/x 1 x2 1 x2 1 x 1 x ln 1
[0, 500] by [0, 3] lim
x0 As x , 7. 4x 2 1 approaches 2. x 1 1 x lim
x0 lim
x0 1 1 x 1
1 x x0 x lim 0 [ 5, 5] by [ 1, 4] Therefore, lim 1
x0 sin 3x As x0, approaches 3. x 1 x x lim f (x)
x0 lim e ln f (x)
x0 e0 1. 8. 4. [0, ] by [ 1, 2] [0, 1000] by [0, 1] As 9. y 10. y tan , 2 2 tan 1 sin h h approaches 1. From the graph, the limit appears to be about 0.714.
5x 2 2 x 7x lim 3x 1 lim
x 10x 3 14x lim
x 10 14 3 11 5 7 0.71429 (1 h)1/h 5. lim x3 1 4x 3 x 3 x1 3x 2 12x 2 1 x1 lim Section 8.1 Exercises
1. [0, 2] by [0, 1] [0, 2] by [0, 1] From the graph, the limit appears to be .
x 2 x2 x 1 4 The graph supports the answer. 6. lim
x0 1 lim 2 4 1 x2 2x lim 1 4 cos x x2 lim
x0 sin x 2x lim
x0 cos x 2 1 2 2. [ 2, 2] by [ 2, 6] [ 5, 5] by [ 1, 1] From the graph, the limit appears to be 5. lim
x0 The graph supports the answer. sin 5x x lim
x0 5 cos 5x 1 5 ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN

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