Unformatted text preview: MAT 319/320 Solutions for HW7 Exercise 1. Section 4.1, #2. √ x − 4 1 x − 2 = √x + 2 x − 4 , therefore if you take x − 4 < 1 you will get Proof. Notice that 2 √ x − 2 < 1/2 and for the second inequality it is suﬃcient to take x − 4 < 2.10− 2. that Exercise 2. Section 4.1, #9d. Proof. One has
x2 − x + 1 x+1 −2 = 1 2x2 − x + 1 − x − 1 2x + 2 = 2x.(x − 1) 2x + 2 = 2x 2x + 2 Now, on the neighborhood [0, + ∞) of the point 1, we have that given ε > 0, for any x ∈ [0, + ∞) satisfying x − 1 < ε one has that Exercise 3. Section 4.1, #10b. Proof. One has
x+5 2x + 3 2x 2x + 2 x2 − x + 1 x+1 . x−1 . 1. Therefore for a − 2 < ε.
1 −4 = x + 5 − 8x − 12 2x + 3 = 2x + 3 . x + 1 is bounded in a neighborhood of − 1. (Notice 7 Now it will be suﬃcient to prove that that it is NOT bounded everywhere! More precisely that function is large when you are too close to − 3/2). Let’s consider the following neighborhood of − 1 given by V = ( − 5/4, 0). 1 −5 Since (x > − 5/4) ⇒ 2x + 3 > 2 + 3 = 2 we get the following inequality: 7 ε on V, 0 < 2x + 3 < 2.7 = 14. Therefore, for a given ε > 0, take δ = 14 . Then for any x ∈ V satisfying x + 1 < 14 , one has
ε x+5 2x + 3 7 2x + 3 − 4 < ε. Thus limx→−1 f (x) = 4. Exercise 4. Section 4.1, #11c. Proof. The functions x x + sgn(x) has a lefthand limit at zero equal to − 1 and a righthand limit equal to + 1, therefore it has no limit at zero. Exercise 5. Section 4.1, #13. Proof. When x → c, to say that f (x)2 → 0 implies that f (x) → 0 (square root rule) which is equivalent to the following phrase: limx→ c f (x) = 0. Now take c = 0, and the function f (x) = signe(x), and declare by convention that f (0) = 1. The square of this function is a constant function equal to 1, but the function itself has no limit at 0. 1 Exercise 6. Section 4.2, #1d. Proof. The sum rule implies that the numerator has a limit equal to 1, and by the sum and product rules the denominator has a limit equal to 2. Then the Quotient rule implies that the limit is 1/2.(Notice that the denominator is never zero!) Exercise 7. Section 4.2, #4. Proof. We proved in class the nonexistence of limx→ 0 cos(1/x) (take the two sequences con1 1 verging to zero given by xn = 2nπ , yn = (2n + 1)π the function takes constant values equal to 1 on the ﬁrst one, and constant values equal to 1 on the other). Now the function x.cos(1/x) has a limit equal to zero at zero because of the squeeze theorem applied to the following inequality: 0 x.cos(1/x) x  → 0, when x → 0. Exercise 8. Section 4.2, #5. Proof. On a neighborhood V of c one has an inequality of the form 0  f (x).g (x) M . g (x), where M is an upper bound for f on V . Now apply the squeze theorem to that inequality to get the result. 2 ...
View
Full Document
 Fall '10
 Mark
 Derivative

Click to edit the document details