Proof of a Limit Value Using Epsilon and Delta — Transcript

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1. 0:00Let's prove the value of a limit using
2. 0:03epsilon delta.
3. 0:11Proof of a limit value
4. 0:16using
5. 0:18It's the definition of a limit. So,
6. 0:19using epsilon
7. 0:23and delta.
8. 0:26Okay?
9. 0:29So, what we want to do and I'll come up
10. 0:32with an example here. Let's say we want
11. 0:33to prove that
12. 0:37prove that
13. 0:39the limit as X approaches three
14. 0:44of the function -2x + 1
15. 0:48= -5.
16. 0:50Okay? We know that this is true. The
17. 0:52limit as X approaches three of this
18. 0:54basic linear function is -5, but how do
19. 0:56we prove it? Well, um what we So, what
20. 1:00we want to prove Let's Let's get a
21. 1:02handle on what we want to prove.
22. 1:05We want
23. 1:08to find
24. 1:10delta
25. 1:12so that
26. 1:16We want to find delta so that when
27. 1:19the distance between X and three
28. 1:22is less than that delta
29. 1:25the distance between f of x, sorry, the
30. 1:28in this case
31. 1:30f of x is -2x + 1
32. 1:33the distance between that and
33. 1:36-5
34. 1:38is less than epsilon for any epsilon.
35. 1:41So, epsilon is given. We need to figure
36. 1:43out delta.
37. 1:45Okay? So, this is our preliminary work.
38. 1:48We need to We need to find delta. We're
39. 1:50going to do that first. Once we find
40. 1:52delta
41. 1:54then we're going to go back and do the
42. 1:55proof. Okay? So, let's find delta.
43. 1:58If I don't know delta,
44. 2:00but I want this epsilon statement to be
45. 2:02true,
46. 2:03let's start here.
47. 2:06Let's start with this and see if this
48. 2:08can give us a hint as to what to choose
49. 2:10delta to be. So, let's let's take the
50. 2:13absolute value of -2x + 1 f of x
51. 2:18L, the limit L is -5.
52. 2:22I want that to be less than epsilon.
53. 2:24Okay?
54. 2:25Let's clean it up, see what happens,
55. 2:28and our goal is basically going to be
56. 2:31can I turn the absolute value into
57. 2:34the absolute value of x - 3. Because if
58. 2:36I can get x - the absolute value of x -
59. 2:393 on the left side, whatever's on the
60. 2:41right, that's going to be what I'm going
61. 2:42to choose for my delta.
62. 2:44So,
63. 2:45-2x + 1
64. 2:47- -5 is the same thing as plus positive
65. 2:50five.
66. 2:51I want that less than epsilon.
67. 2:54So, I've got -2x + 1 + 5, which is plus
68. 2:57six,
69. 2:59is less than epsilon.
70. 3:02And then I look inside the absolute
71. 3:04value and I compare it to what I want. I
72. 3:06want this. In particular, I want x - 3
73. 3:10in the absolute value.
74. 3:12So, I notice what I can do here is I can
75. 3:14factor this -2 out.
76. 3:17And that leaves positive x and -3.
77. 3:21Again, that's inside the absolute value.
78. 3:24Now, the absolute value
79. 3:28of a negative or a positive is the same.
80. 3:32So, what I can do here is I can say,
81. 3:34"Well,
82. 3:35this is the same thing as the absolute
83. 3:37value of positive two times x - 3." It
84. 3:40doesn't matter,
85. 3:41right? Because a negative times this is
86. 3:43going to be one value, and when you take
87. 3:45the absolute value, it's going to be
88. 3:46positive. Or you can do positive two
89. 3:48times this and take the absolute value
90. 3:50and still get a positive.
91. 3:51Okay?
92. 3:52And now
93. 3:55the only barrier between this absolute
94. 3:56value and what I want is this two. But
95. 3:59since the two is positive, it's the same
96. 4:02thing as multiplying by positive two
97. 4:04outside
98. 4:05of the absolute value. And that's the
99. 4:07key.
100. 4:08How do you factor and pull the constant
101. 4:11out so that what remains is what you
102. 4:13want?
103. 4:14So, you have to you have to understand
104. 4:16absolute values well to understand you
105. 4:18can lose the negative. And then because
106. 4:20this is positive, you might as well just
107. 4:21multiply by the positive two outside the
108. 4:23absolute value.
109. 4:25And now, to get the absolute value by
110. 4:26itself, it's very simple. Divide both
111. 4:29sides by two.
112. 4:35So, what we're going to do is now this
113. 4:39this statement is equivalent to this
114. 4:42epsilon statement we started with
115. 4:43because it we we did it through a series
116. 4:46of equivalent statements.
117. 4:48And so, we know that if this is true,
118. 4:50that's going to be true. So, what we're
119. 4:52going to do is we're going to choose
120. 4:53this to be the delta in our proof. So,
121. 4:57delta equals epsilon over two. Let's
122. 4:59remember that.
123. 5:10So, here goes the proof finally. We've
124. 5:11done our preliminary work. We know our
125. 5:12delta.
126. 5:14So, our proof is going to look like
127. 5:15this.
128. 5:20Suppose
129. 5:21right? How does the definition start?
130. 5:23Well, you have to be given epsilon.
131. 5:26So, suppose epsilon greater than zero is
132. 5:28given.
133. 5:31Okay? Suppose epsilon greater than zero
134. 5:33is given.
135. 5:37Choose
136. 5:40delta equal to epsilon over two, like we
137. 5:44saw before.
138. 5:46Okay?
139. 5:53And what we're going to do is
140. 5:55now
141. 5:56the definition says that if the absolute
142. 6:00value of x minus three
143. 6:03is less than delta, which is epsilon
144. 6:07over two.
145. 6:09If that's true, then the absolute value
146. 6:11of -2x + 1 - -5 should be less than
147. 6:15epsilon, right? So, that's what we're
148. 6:17going to look at now. So, assume this is
149. 6:19true.
150. 6:22We want to because of this, we want to
151. 6:24be able to state that the absolute value
152. 6:26of this function minus -5 is
153. 6:29uh less than epsilon. So, let's examine
154. 6:34So, now the absolute value of -2x + 1
155. 6:39minus -5
156. 6:43equals the absolute value
157. 6:46of -2
158. 6:49x + 1
159. 6:51+ 5.
160. 6:52I know this work is familiar. We had to
161. 6:54do a lot of the same work in our
162. 6:55preliminary setup to find our delta.
163. 7:00Equals
164. 7:01the absolute value of -2x + 6,
165. 7:05which is the same thing as the absolute
166. 7:06value of -2 * the quantity x - 3.
167. 7:10Let's come over here.
168. 7:14If I had regular notebook paper, I'd
169. 7:16work vertically.
170. 7:17Equals the absolute value of 2 * x - 3,
171. 7:23which equals pull the two out, 2 * the
172. 7:26absolute value of x - 3, and now this is
173. 7:29where it's different.
174. 7:30What do we know about the absolute value
175. 7:32of x - 3?
176. 7:34The absolute value of x - 3 is less than
177. 7:37epsilon over two.
178. 7:39So, that means if I take 2 * this
179. 7:41positive number, it's going to be less
180. 7:43than two times
181. 7:46Well, the absolute value of x minus
182. 7:47three is less than epsilon over two. So,
183. 7:50this this product is going to be less
184. 7:52than the product of two and epsilon over
185. 7:55two, which equals epsilon.
186. 7:59Okay? So, let's let's go back.
187. 8:02We assumed the delta statement.
188. 8:06We assumed that the absolute value of x
189. 8:07minus three is less than this delta
190. 8:09epsilon over two.
191. 8:12And we concluded
192. 8:15Thus,
193. 8:16let's put it all together now. Thus,
194. 8:19this guy here, the absolute value
195. 8:23of -2x + 1 - -5
196. 8:27equals this, equals this, equals this,
197. 8:30equals this, equals this, and finally is
198. 8:33less than this, which equals epsilon.
199. 8:36So, in other words, the original
200. 8:39expression
201. 8:40is less than
202. 8:43epsilon.
203. 8:44And that's the end of the proof.
204. 8:48Because we proved that if we assume this
205. 8:50is true,
206. 8:52then that let us to conclude that this
207. 8:54is true, and that's what the definition
208. 8:57of the limit says has to be true.
209. 8:59If this distance is less than the delta,
210. 9:02then this distance is less than epsilon,
211. 9:04and it's true in this case as long as we
212. 9:06pick delta to be epsilon over two, or
213. 9:08anything smaller than that. Okay? So,
214. 9:11this is a proof
215. 9:13of a limit using epsilon and delta.