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Let fbe a continuous function from R to R Prove that fx f(x) = 0gis a closed subset of R Solution Let y be a limit point of fx f(x) = 0g So there is a sequence fy ngsuch that y n 2fx f(x) = 0gfor all nand lim n!1y n = y Since f is continuous, by Theorem 402 we have f(y) = lim n!1f(y. F r o m t h i s e x p l a n a t i o n y o u s h o u l d b e a b l e t o f a c t or t h e e x p r e ss i o n s i n E x e r c i s e 1 1 0 D o n. Ϗd ˂ ܂܎ o P X Carico OREA L R I A 798 ~ i ō j y X ܎ u z R t P X 74 N A 40× s 74× 31cm.

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@ ɗ ΁A C X G A ǂ Ȃ Ƃ Ă S n Ƃ ł Ȃ R ܂ B f l ̖ڂł A S ɐ헪 I ȓy n ł 邱 Ƃ ł 傤 B ́A ܂ A ̑ ̓y n Ɋւ A C X G Ǝ ӃC X ̖ ɂ Ă @ ^ ܂ B ܂ A u a ̂ ߂̓y n v Ƃ _ ͒ʗp Ȃ Ƃł B ̖K n K ɍs ΁A j ̂ Ƃ𕨌 Ă 邱 Ƃ m ̂ł A ̃S ɓ ΁A K ΂ ̎x z ɓ B K ΂ ̎x z ɓ ΁A ͋M d Ȑ A ̃C X G S ̂ x z ɓ 邱 Ƃ ł A Ƃ ƂɂȂ. Proof lnexy = xy = lnex lney = ln(ex ·ey) Since lnx is onetoone, then exy = ex ·ey 1 = e0 = ex(−x) = ex ·e−x ⇒ e−x = 1 ex ex−y = ex(−y) = ex ·e−y = ex · 1 ey ex ey • For r = m ∈ N, emx = e z }m { x···x = z }m { ex ···ex = (ex)m • For r = 1 n, n ∈ N and n 6= 0, ex = e n n x = e 1 nx n ⇒ e n x = (ex) 1 • For r rational, let r = m n, m, n ∈ N. G ^_ ^ h _ g b N_ G A T X F t @ C X g _ Ԏ _ G A T X_ L b g y X c C g F s t m Q Q P Q z ԍ t g A b v A ␳ w p G A T X ł B H Ŏ t ł u { g I d l v A x ̉.

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