Chapter 2 | Limits
201
taking a look at our ultimate desired inequality: | ⎛
⎠ −6 | < ε . This inequality is equivalent
⎝ x 2 −2 x +3 ⎞
2.29 x 2 =1. Let ε >0; choose δ =min{1, ε /3}; assume 0< | x −1| < δ . Since | x −1| <1, we may conclude that −1< x −1<1. Thus, 1< x +1<3. Hence, | x +1 | <3. Complete the proof that lim x →1 You will find that, in general, the more complex a function, the more likely it is that the algebraic approach is the easiest to apply. The algebraic approach is also more useful in proving statements about limits. Proving Limit Laws We now demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws. The triangle inequality is used at a key point of the proof, so we first review this key property of absolute value. to | x +1 | · | x −3| < ε . At this point, the temptation simply to choose δ = ε x −3 is very strong. Unfortunately, our choice of δ must depend on ε only and no other variable. If we can replace | x −3| by a numerical value, our problem can be resolved. This is the place where assuming δ ≤1 comes into play. The choice of δ ≤1 here is arbitrary. We could have just as easily used any other positive number. In some proofs, greater care in this choice may be necessary. Now, since δ ≤1 and | x +1 | < δ ≤1, we are able to show that | x −3| <5. Consequently, | x +1 | · | x −3| < | x +1 | ·5. At this point we realize that we also need δ ≤ ε /5. Thus, we choose δ =min{1, ε /5}. 3. Assume 0< | x +1 | < δ . Thus, | x +1 | <1and | x +1 | < ε 5 . Since | x +1 | <1, we may conclude that −1< x +1<1. Thus, by subtracting 4 from all parts of the inequality, we obtain −5< x −3<−1. Consequently, | x −3| <5. This gives us | ⎛ ⎝ x 2 −2 x +3 ⎞ ⎠ −6 | = | x +1 | · | x −3| < ε 5 ·5= ε . Therefore, lim x →−1 ⎛ ⎝ x 2 −2 x +3 ⎞ ⎠ =6.
Definition The triangle inequality states that if a and b are any real numbers, then | a + b | ≤ | a | + | b | .
Proof We prove the following limit law: If lim x → a
g ( x ) = M , then lim x → a ⎛
⎝ f ( x )+ g ( x ) ⎞
f ( x ) = L and lim x → a
⎠ = L + M .
Let ε >0. Choose δ 1 >0 so that if 0< | x − a | < δ 1 , then | f ( x )− L | < ε /2. Choose δ 2 >0 so that if 0< | x − a | < δ 2 , then | g ( x )− M | < ε /2.
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