Advanced Calculus (II)
WEN-CHINGLIEN
Department of Mathematics National Cheng Kung University
2009
Ch9: Convergence in R
9.2: Limits Of Functions
Definition (9.13)
Let n, m ∈N and a ∈ Rn,let V be an open set which containsa, and suppose that f : V \{a} → Rm. Then f (x) is said to converge toL, as x approaches a, if and only if for every ε > 0 there is a δ > 0 (which in general depends on ε, f , V , anda) such that
0 < kx − ak ≤ δ implies kf (x) − Lk < ε.
In this case we write
L = lim
x→af (x)
and callL the limit of f(x) as x approaches a.
Let n, m ∈N, let a ∈ Rn,let V be an open ball which containsa, and let f , g : V \{a} → Rm.
(i) If f (x) = g(x) for all x ∈ V \{a} and f (x) has a limit as x → a, then g(x) also has a limit as x → a, and
x→alimg(x) = lim
x→af (x).
(ii) [Sequential Characterization of Limits]
L = limx→af (x) exists if and only if f (xk) →L as k → ∞ for every sequencexk ∈ V \{a} which converges to a as k → ∞.
(iii) Suppose that α ∈R. If f (x) and g(x) have limits, as x approachesa, then so do (f + g)(x), (αf )(x), (f · g)(x), and kf (x)k. In fact,
Let n, m ∈N, let a ∈ Rn,let V be an open ball which containsa, and let f , g : V \{a} → Rm.
(i) If f (x) = g(x) for all x ∈ V \{a} and f (x) has a limit as x → a, then g(x) also has a limit as x → a, and
x→alimg(x) = lim
x→af (x).
(ii) [Sequential Characterization of Limits]
L = limx→af (x) exists if and only if f (xk) →L as k → ∞ for every sequencexk ∈ V \{a} which converges to a as k → ∞.
(iii) Suppose that α ∈R. If f (x) and g(x) have limits, as x approachesa, then so do (f + g)(x), (αf )(x), (f · g)(x), and kf (x)k. In fact,
Let n, m ∈N, let a ∈ Rn,let V be an open ball which containsa, and let f , g : V \{a} → Rm.
(i) If f (x) = g(x) for all x ∈ V \{a} and f (x) has a limit as x → a, then g(x) also has a limit as x → a, and
x→alimg(x) = lim
x→af (x).
(ii) [Sequential Characterization of Limits]
L = limx→af (x) exists if and only if f (xk) →L as k → ∞ for every sequencexk ∈ V \{a} which converges to a as k → ∞.
(iii) Suppose that α ∈R. If f (x) and g(x) have limits, as x approachesa, then so do (f + g)(x), (αf )(x), (f · g)(x), and kf (x)k. In fact,
x→alim(f + g)(x) = lim
x→af (x) + lim
x→ag(x),
x→alim(αf )(x) = α lim
x→af (x),
x→alim(f · g)(x) = lim
x→af (x) · lim
x→ag(x),
lim
x→af (x) = lim
x→akf (x)k . Moreover, when m = 3,
x→alim(f × g)(x) = lim
x→af (x) × lim
x→ag(x), and when m = 1 and the limit of g is nonzero,
x→alimf (x)/g(x) = lim
x→af (x)/ lim
x→ag(x).
(iv) [Squeeze Theorem for Functions] Suppose that f , g, h : V \{a} → R and g(x) ≤ h(x) ≤ f (x) for all x ∈ V \{a}. If
x→alimf (x) = lim
x→ag(x) = L, then the limit of h also exists, asx → a, and
x→alimh(x) = L.
(v) Suppose that U is open inRm, thatL ∈ U, and h : U\{L} → Rpfor some p ∈N. If L = limx→ag(x) and M = limy→Lh(y), then
x→alimh ◦ g(x) = h(L).
(iv) [Squeeze Theorem for Functions] Suppose that f , g, h : V \{a} → R and g(x) ≤ h(x) ≤ f (x) for all x ∈ V \{a}. If
x→alimf (x) = lim
x→ag(x) = L, then the limit of h also exists, asx → a, and
x→alimh(x) = L.
(v) Suppose that U is open inRm, thatL ∈ U, and h : U\{L} → Rpfor some p ∈N. If L = limx→ag(x) and M = limy→Lh(y), then
x→alimh ◦ g(x) = h(L).
Leta ∈ Rn,let V be an open ball that containsa, let f = (f1, . . . ,fm) :V \{a} → Rm,and let
L = (L1,L2, . . . ,Lm) ∈Rm.Then
(1) L = lim
x→af (x) exists if and only if
(2) Lj = lim
x→afj(x) exists for each j = 1, 2, . . . , m.
Prove that
f (x , y ) = 3x2y x2+y2 converges as (x , y ) → (0, 0).
Prove that the function
f (x , y ) = 2xy x2+y2 has no limit as (x , y ) → (0, 0).
Evaluate the iterated limits of f (x , y ) = x2
x2+y2 at (0,0)
Sol. (9.20)
For each x 6= 0, x2x+y2 2 → 1 as y → 0. Therefore,
x →0lim lim
y →0
x2
x2+y2 = lim
x →0
x2 x2 =1.
On the other hand,
y →0lim lim
x →0
x2
x2+y2 = lim
y →0
02 y2 =0
Suppose that I and J are open intervals, that a ∈ I and b ∈ J, and that f : (I × J)\{(a, b)} →R. If
g(x ) := lim
y →bf (x , y )
exists for each x ∈ I\{a}, if limx →af (x , y ) exists for each y ∈ J\{b}, and if f (x , y ) → L as (x , y ) → (a, b) (inR2), then
L = lim
x →alim
y →bf (x , y ) = lim
y →blim
x →af (x , y ).
