函数极限性。简单而言,上一节定义的limiting values of functions如果等于函数在这一点的值,就是函数连续。连续性用ε-δ定义和序列定义是等价的,可以证明很多函数都是连续的
Exercise 9.4.1
( a ) ⇔ ( b ): (f is continuous at x0)⇔(x→x0;x∈Xlimf(x)=f(x0))⇔((∀(an)n=0∞∈E,n→∞liman=x0)⇒n→∞limf(an)=f(x0)) ( a ) implies ( c ): Get directly from the definition of limx→x0;x∈Xf(x)=f(x0). ( c ) implies ( d ): Obviously true. ( d ) implies ( a ): Let ∀ϵ>0, then due to (d), ∃δ′>0, s.t. ∣f(x)−f(x0)∣≤ϵ/2 for every x∈X with ∣x−x0∣≤δ′, so if we let δ=δ′, then ∣x−x0∣<δ means ∣x−x0∣≤δ′, and ∣f(x)−f(x0)∣≤ϵ/2⇒∣f(x)−f(x0)∣<ϵ⇒x→x0;x∈Xlimf(x)=f(x0)
Exercise 9.4.2
First for every x0∈X, we have x→x0;x∈Xlimf(x)=x→x0;x∈Xlimc=c=f(x0) So f is continuous on X. Next, for every x0∈X, we have x→x0;x∈Xlimg(x)=x→x0;x∈Xlimx=x0=g(x0) So g is continuous on X.
Exercise 9.4.3
Choose ∀x0∈R, we prove f(x)=ax is continuous at x0. As a>0, we have ax0>0. ∀ϵ>0, from Lemma 6.5.3 we can find a N∈N, s.t. ∣a1/n−1∣<ϵ/ax0 for all n>N. Since ∣ax−ax0∣=ax0∣ax−x0−1∣ We choose δ′=1/(N+1), then ∀x∈R with 0≤x−x0<δ′, we can find a k>N, s.t. 1/(k+1)<x−x0<1/k Thus ∣ax−x0−1∣ is between ∣a1/k−1∣ and ∣a1/k+1−1∣, which means ∣ax−x0−1∣<ϵ/ax0⇒∣ax−ax0∣=ax0∣ax−x0−1∣<ϵ Also, from Lemma 6.5.3 we can find a N′∈N, s.t. ∣a−1/n−1∣<ϵ/ax0 for all n>N′. Since ∣ax−ax0∣=ax0∣ax−x0−1∣ We choose δ′′=1/(N′+1), then ∀x∈R with 0<x0−x<δ′′, we can find a k>N′, s.t. 1/(k+1)<x0−x<1/k Thus ∣ax−x0−1∣ is between ∣a−1/k−1∣ and ∣a−1/k+1−1∣, which means ∣ax−x0−1∣<ϵ/ax0⇒∣ax−ax0∣=ax0∣ax−x0−1∣<ϵ Now let δ=min(δ′,δ′′), then if ∣x−x0∣<δ,∣ax−ax0∣<ϵ, completing the proof.
Exercise 9.4.4
Since limx→1x=1, we have limx→1xn=(limx→1x)n=1 for all natural numbers n, and since when x→1,1/x→1, so limx→1x−n=1,n∈N. Use prove by contradiction we can show that limx→1x1/n=1,∀n∈N+, thus it’s able to conclude limx→1xq=1,∀q∈Q. Now for ∀p∈R,∃q1,q2,s.t.q1≤p<q2, so xp is between xq1 and xq2, which means x→1limxp=1,∀p∈R Now for x0∈(0,+∞), then x0p>0. For ∀ϵ>0,∃δ>0, s.t. for any x∈(0,+∞) with ∣x−x0∣<δ, we have ∣(x/x0)p−1∣<ϵ/x0p, thus ∣xp−x0p∣=x0p∣(x/x0)p−1∣<ϵ
Exercise 9.4.5
Let ∀ϵ>0, as g is continuous at f(x0), we can find δ1>0, s.t. ∀y∈Y,∣y−f(x0)∣<δ1, we can have ∣g(y)−g(f(x0))∣<ϵ. For this δ1>0 we can find δ>0, s.t. ∀x∈X,∣x−x0∣<δ, we can have ∣f(x)−f(x0)∣<δ1, thus if ∣x−x0∣<δ, we get ∣g(f(x))−g(f(x0))∣<ϵ⇒∣(g∘f)(x)−(g∘f)(x0)∣<ϵ
Exercise 9.4.6
Let y0∈Y, so y0∈X, and f is continuous at y0, so ∀ϵ>0,∃δ>0, s.t. ∀x∈X,∣x−y0∣<δ, we can have ∣f(x)−f(y0)∣<ϵ. Use the result above, if y∈Y,∣y−y0∣<δ, we can have ∣f(y)−f(y0)∣<ϵ, this proves f∣Y:Y→R is continuous at y0, since y0 is arbitrarily chosen, f∣Y is continuous.
Exercise 9.4.7
By Proposition 9.4.9, the function xi is continuous for each 0≤i≤n, thus the function cixi is continuous for each 0≤i≤n, at last their sums is continuous.