8. LimitsDefinition 8.1. Let P ∈ Rn be a point. The open ball of radius �> 0 about P is the set B�(P )= { Q ∈ Rn |�−→PQ� <� }. The closed ball of radius �> 0 about P is the set { Q ∈ Rn �−→| PQ�≤ � }. 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 ∈ N }⊂ R. 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, QP→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 functioons Let λ ∈ R be a scalar. If P is a limit point of A and lim f(Q)= L and lim g(Q)= M, QP QP→→then (1) limQP (f + g)(Q)= L + M, and →(2) limQP (λf)(Q)= λL. →Now suppose that m =1. (3) limQP (fg)(Q)= LM, and →(4) if M =0, then limQP (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 ). QP→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 + y2 is continuous. Sometimes Larry is very lucky: 2 Example 8.12. Does the limit 22 lim x− y, (x,y)(0,0)→x − y 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 22x− y= x + y, x − y so that 22 lim x− y= lim x + y =0. (x,y)(0,0) (x,y)(0,0)→x − y →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 Q→P f(Q) = L and lim M→L g(M) = E. Then lim Q→P (g ◦ f)(Q) = E. 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 �(gf)(Q) − E� = �g(f(Q)) − E�◦ = �g(M) − E� < �. � Example 8.14. Does xylim (x,y)→(0,0) x2 + 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 =lim 0=0. t0 t2 +0 t0→→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 11 lim =lim = . t0 t2 + t2 t0 22→→The problem is that the limit along two different lines is different. So the original limit cannot exist. Example 8.15. Does the limit 3xlim , (x,y)→(0,0) x2 + y2 exist? Let us use polar coordinates. Note that x3 r3 cos3 θ == r cos 3 θ. x2 + y2 r2 So we guess the limit is zero. 3 lim x= lim r cos 3 θ(x,y)→(0,0) |x2 + y2 | r→0 || ≤ lim |r| =0. r0→Example 8.16. Does the limit xyz lim , (x,y,z)→(0,0,0) x2 + y2 + z2 exist? Same trick, but now let us use spherical coordinates. xyz ρ3 cos2 φ sin φ cos θ sin θ lim = lim (x,y,z)→(0,0,0) |x2 + y2 + z2 | ρ→0 | ρ2 | = lim ρ cos 2 φ sin φ cos θ sin θρ→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 ylim , (x,y)→(0,0) y + x2 exist? If we restrict to the line t −→ (at,bt), then we get bt b lim =lim =1. t→0 bt + a2t2 t→0 b + at But if we restrict to the conic t −→ (t,at2), then we get at2 aa lim =lim = , t0 at2 + t2 t0 1+ a 1+ a→→and the limit changes as we vary a, so that the limit does not exist. Note that if we start with y y + xd , then Norman even needs to use curves of degree d, t −→ (t,atd). 5 MIT OpenCourseWarehttp://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.