Download Lecture Notes on Exercise 36 - Calculus and Analytic Geometry I | MATH 109 and more Study notes Analytical Geometry and Calculus in PDF only on Docsity! A note on exercise #36 in Section 2.4 of Stewart’s Calculus Use the ε,δ definition of limit (on page 115) to prove that: limit x→2 1/x = ½ . What we need to do here is to show how, for any given number ε > 0 we can select some corresponding number δ > 0 so that, if x is chosen to satisfy | x – 2 | < δ , then the function-value 1/x will satisfy | 1/x – 1/2 | < ε . After doing some scratch work, we have discovered how the choice for δ can be made, so we can now present our proof. Proof: Suppose ε > 0 has been selected. Let us then choose δ = minimum{ 1, 2ε }. This choice for δ will assure us that 0 < δ ≤ 1 and also that δ ≤ 2ε , two facts which we will need to use in the reasoning below. So let us now assume that x is any number which satisfies the inequality | x – 2 | < δ . We know we can multiply both sides of this inequality by any positive value and retain the direction of the inequality sign. In particular, we could multiply by 1/| 2x | if we knew that x was nonzero. But because of our prior assumption that | x – 2 | < δ and because we chose δ ≤ 1 , we can be sure that x lies in the interval 1 ≤ x ≤ 3 , and so x cannot be zero. Therefore by multiplication we can obtain the inequality: | x – 2 |/| 2x | < δ/| 2x | . Remembering (from high school algebra) that a quotient of absolute values equals the absolute value of the quotient, and remembering the rules for adding two fractions, we see that we can rewrite the left-hand term of this inequality: | x – 2 |/| 2x | = | (x - 2)/2x | = | 1/2 – 1/x | = | 1/x – 1/2 | Thus our inequality becomes: | 1/x – 1/2 | < δ/| 2x | , and, since x is larger than 1, we can drop the absolute-value sign on the right side: | 1/x – 1/2 | < δ/ 2x . But if x is larger than 1, then 1/2x will be smaller than ½ , so we get: | 1/x – 1/2 | < δ/ 2x < δ/ 2 . Remembering that we chose δ so that δ ≤ 2ε , we see that we can get: | 1/x – 1/2 | < δ/ 2 ≤ 2ε/2 = ε . As we explained at the outset, this was all that we needed to show! ■
