University of California, Berkeley

Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
Problem 61
Problem 62
Problem 63
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Problem 71
Problem 72
Problem 73
Problem 74
Problem 75
Problem 76
Problem 77
Problem 78
Problem 79
Problem 80
Problem 81
Problem 82
Problem 83
Problem 84
Problem 85
Problem 86

Problem 36

Finding a $\delta$ for a Given $\varepsilon$ The graph of

$f(x)=\frac{1}{x-1}$

is shown in the figure. Find $\delta$ such that if $0<|x-2|<\delta$ then $|f(x)-1|<0.01$

Answer

$\delta=\frac{1}{101}$

You must be logged in to like a video.

You must be logged in to bookmark a video.

## Discussion

## Video Transcript

okay from this from the Graf Spee feet out up to is equal to one. So the function might the limits. One is less than zero point there. One which means at one point there one lesson. Actually, that's the wrong way. Could be no 0.99 left in the back. Just one point 1.1. Okay. From the figure we see that effort Valley did that to you? Their one over 101 Is he going to 1.1? And this is all we waited at 199 over 99.99 So when ex doctor value of X minus two is last in Delta, that's that for a function minus the limit. Is that the 0.1? It gives us that to a one over 101 in less than X. What? You left in 199 over 99. Therefore, our delta is equal to the minimum of two. When it's to their one older one over one or a 9199 over 99 2 And the minimum is one over one on one

## Recommended Questions