0 0 such that f x L whenever x a

Transcription

1 Epsilon-Delta Definition of the Limit Few statements in elementary mathematics appear as cryptic as the one defining the limit of a function f() at the point = a, 0 0 such that f L whenever a Translation: for every epsilon greater than zero, there eists a delta greater than zero such that f() lies within epsilon units of the limit L whenever lies within delta units of a l L, f y=l+ y=l- a- a a+ With the aid of a graphic interpretation, the definition is less intimidating. The definition says that the limit of f() at =a is L if and only if an interval of the domain around a can be found for every interval of the range around L. In other words, to defend a limit result, we must produce a value for in response to the challenge of any. We can think of the epsilon-delta definition as a game. The game is played as follows. We are challenged with an arbitrary positive number and must then produce a corresponding positive number that does "appropriate things". We usually think of as small (if we can find 's for small 's, then we can find them for larger ones) The, process can be visualized as a game of lines. If you are given two horizontal lines y=l+ and y=l-, you must then produce two vertical lines at a and a, so that the rectangle formed by the four lines has the property that the graph stays in the rectangle for all 's between a- and a+, ecept possibly at =a itself. You get to choose >0. lim f L a In proving the limit, we must start with a and use algebra to show that this is equivalent to and. f L. The challenge to the proof is finding the relationship between

2 The best way to find the (that will depend on any ) is to actually start with f L then use algebra to transform it into a which will provide the relationship between and. Eample Prove lim 1 7 In order to prove any limit, we must use the, definition. That is, given and number 0, we must find a number 0 Step 1: To find a (depending on ) such that Start with 1 7 and transform into 1 7 whenever Therefore. Step : The formal proof. Given any 0, we choose 0 to be Therefore 1 7 whenever, as required.

3 Eample Prove the following using epsilons and deltas lim 1 5 In order to prove any limit, we must use the, definition. That is, given and number 0, we must find a number 0 Step 1: To find a (depending on ) such that Start with 1 5 and transform into 1 5 whenever Therefore. Step : The formal proof. Given any 0, we choose 0 to be Therefore 1 5 whenever, as required.

4 Eample Prove lim 5 17 In order to prove any limit, we must use the, definition. That is, given and number 0, we must find a number 0 Step 1: To find a (depending on ) such that Start with 5 17 and transform into 5 17 whenever Notice how we have two absolute value terms to deal with (don t worry). This function requires a two-step approach to find. First choose the term that matches -a, and let that term be less than a easy number to work with (choose a small number like or even better 1). We will then use algebra to transform the first term into the second term to find the inequality relationship of that second term. We choose 1 so that: Therefore implies , since Thus we have two conditions on : 1 and. So we choose to be the 15 smaller of 1 and. Then 0 implies

5 Eample Prove lim 9 In order to prove any limit, we must use the, definition. That is, given and number 0, we must find a number 0 such that Step 1: To find a (depending on ) Start with 9 and transform into 9 9 whenever. Notice how we have two absolute value terms to deal with (don t worry). This function requires a two-step approach to find. First choose the term that matches -a, and let that term be less than a easy number to work with (choose a small number like or even better 1). We will then use algebra to transform the first term into the second term to find the inequality relationship of that second term. We choose 1 so that Therefore 0 1 implies 9 7 ;since 7 7 Thus we have two conditions on : 1 and. So we choose to be the 7 smaller of 1 and. Then 0 implies 9. 7

6 Eample Prove lim In order to prove any limit, we must use the, definition. That is, given and number 0, we must find a number 0 such that 1 1 whenever. Step 1: To find a (depending on ) Start with 1 1 and transform into Notice how we have two absolute value terms to deal with (don t worry). This function requires a two-step approach to find. First choose the term that matches -a, and let that term be less than a easy number to work with (choose a small number like or even better 1). We will then use algebra to transform the first term into the second term to find the inequality relationship of that second term. We choose 1 so that

7 Therefore 0 1 implies: 1 Thus we have two conditions on : 1 and. So we choose to be the smaller of 1 and. Then 0 implies.