Calculus of Several Variables, Lecture 8: Limits

8. Limits Definition 8.1. Let P ∈ Rn be a point. The open ball of radius � > 0 about P is the set −→ B� (P ) = { Q ∈ Rn | �P Q� < � }. The closed ball of radius � > 0 about P is the set −→ { Q ∈ Rn | �P Q� ≤ � }. Definition 8.2. A subset A ⊂ Rn is called open if for every P ∈ A there is an � > 0 such that the open ball of radius � about P is entirely contained in A, B� (P ) ⊂ A. We say that B is closed if the complement of B is open. Put differently, an open set is a union of open balls. Open balls are open and closed balls are closed. [0, 1) is neither open nor closed. Definition 8.3. Let B ⊂ Rn . We say that P ∈ Rn is a limit point if for every � > 0 the intersection B� (P ) ∩ B = � ∅. Example 8.4. 0 is a limit point of 1 { | n ∈ N } ⊂ R. n Lemma 8.5. A subset B ⊂ Rn is closed if and only if B contains all of its limit points. Example 8.6. Rn − {0} is open. One can see this directly from the definition or from the fact that the complement {0} is closed. Definition 8.7. Let A ⊂ Rn and let P ∈ Rn be a limit point. Suppose that f : A −→ Rm is a function. We say that f approaches L as Q approaches P and write lim f (Q) = L, Q→P if for every � > 0 we may find δ > 0 such that whenever Q ∈ Bδ (P ) ∩ A f (Q) ∈ B� (L). In this case we call L the limit. It might help to understand the notion of a limit in terms of a game played between two people. Let’s call the first player Larry and the second player Norman. Larry wants to show that L is the limit of f (Q) as Q approaches P and Norman does not. So Norman gets to choose � > 0. Once Norman has chosen � > 0, Larry has to choose δ > 0. The smaller that Norman chooses � > 0, 1 the harder Larry has to work (typically he will have to make a choice of δ > 0 very small). Proposition 8.8. Let f : A −→ Rm and g : A −→ Rm be two func­ tions. Let λ ∈ R be a scalar. If P is a limit point of A and lim f (Q) = L Q→P and lim g(Q) = M, Q→P then (1) limQ→P (f + g)(Q) = L + M , and (2) limQ→P (λf )(Q) = λL. Now suppose that m = 1. (3) limQ→P (f g)(Q) = LM , and (4) if M �= 0, then limQ→P (f /g)(Q) = L/M . Proof. We just prove (1). Suppose that � > 0. As L and M are limits, we may find δ1 and δ2 such that, if �Q − P � < δ1 and Q ∈ A, then �f (Q) − L� < �/2 and if �Q − P � < δ2 and Q ∈ A, then �g(Q) − L� < �/2. Let δ = min(δ1 , δ2 ). If �Q − P � < δ and Q ∈ A, then �(f + g)(Q) − L − M � = �(f (Q) − L) + (g(Q) − M )� ≤ �(f (Q) − L)� + �(g(Q) − M )� � � ≤ + 2 2 = �, where we applied the triangle inequality to get from the second line to the third line. This is (1). (2-4) have similar proofs. � Definition 8.9. Let A ⊂ Rn and let P ∈ A. If f : A −→ Rm is a function, then we say that f is continuous at P , if lim f (Q) = f (P ). Q→P We say that f is continuous, if it continuous at every point of A. Theorem 8.10. If f : Rn −→ R is a polynomial function, then f is continuous. A similar result holds if f is a rational function (a quotient of two polynomials). Example 8.11. f : R2 −→ R given by f (x, y) = x2 + y 2 is continuous. Sometimes Larry is very lucky: 2 Example 8.12. Does the limit x2 − y 2 , (x,y)→(0,0) x − y lim exist? Here the domain of f is A = { (x, y) ∈ R2 | x = � y }. Note (0, 0) is a limit point of A. Note that if (x, y) ∈ A, then x2 − y 2 = x + y, x−y so that x2 − y 2 lim = lim x + y = 0. (x,y)→(0,0) x − y (x,y)→(0,0) So the limit does exist. Norman likes the following result: Proposition 8.13. Let A ⊂ Rn and let B ⊂ Rm . Let f : A −→ B and g : B −→ Rl . Suppose that P is a limit point of A, L is a limit point of B and lim f (Q) = L and Q→P lim g(M ) = E. M →L Then lim (g ◦ f )(Q) = E. Q→P Proof. Let � > 0. We may find δ > 0 such that if �M − L� < δ, and M ∈ B, then �g(M ) − E� < �. Given δ > 0 we may find η > 0 such that if �Q − P � < η and Q ∈ A, then |f (Q) − L� < η. But then if �Q − P � < η and Q ∈ A, then M = f (Q) ∈ B and �M − L� < δ, so that �(g ◦ f )(Q) − E� = �g(f (Q)) − E� = �g(M ) − E� < �. Example 8.14. Does lim (x,y)→(0,0) x2 � xy + y2 exist? The answer is no. To show that the answer is no, we suppose that the limit exists. Suppose we consider restricting to the x-axis. Let f : R −→ R2 , 3 be given by t −→ (t, 0). As f is continuous, if we compose we must get a function with a limit, 0 lim 2 = lim 0 = 0. t→0 t + 0 t→0 Now suppose that we restrict to the line y = x. Now consider the function f : R −→ R2 , be given by t −→ (t, t). As f is continuous, if we compose we must get a function with a limit, t2 1 1 = lim = . t→0 t2 + t2 t→0 2 2 The problem is that the limit along two different lines is different. So the original limit cannot exist. lim Example 8.15. Does the limit x3 , (x,y)→(0,0) x2 + y 2 lim exist? Let us use polar coordinates. Note that x3 r3 cos3 θ = = r cos3 θ. 2 2 2 x +y r So we guess the limit is zero. | lim (x,y)→(0,0) x3 | = lim |r cos3 θ| r→0 x2 + y 2 ≤ lim |r| = 0. r→0 Example 8.16. Does the limit lim (x,y,z)→(0,0,0) x2 xyz , + y2 + z2 exist? Same trick, but now let us use spherical coordinates. lim (x,y,z)→(0,0,0) | xyz ρ3 cos2 φ sin φ cos θ sin θ | = lim | | ρ→0 x2 + y 2 + z 2 ρ2 = lim |ρ cos2 φ sin φ cos θ sin θ| ρ→0 ≤ lim |ρ| = 0. ρ→0 Sometimes Norman needs to restrict to more complicated curves than just lines: 4 Example 8.17. Does the limit y , (x,y)→(0,0) y + x2 exist? If we restrict to the line t −→ (at, bt), then we get bt b lim = lim = 1. 2 2 t→0 bt + a t t→0 b + at But if we restrict to the conic t −→ (t, at2 ), then we get lim at2 a a = lim = , 2 2 t→0 at + t t→0 1 + a 1+a and the limit changes as we vary a, so that the limit does not exist. lim Note that if we start with y , y + xd then Norman even needs to use curves of degree d, t −→ (t, atd ). 5