monoton funksiyalar uzluksizligi. uzilish nuqtalari va uzluksizli

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1576482411.doc 0 x 0 x e 0 x e £ 0 x £ 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x î 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 2 , b a a + b b a ; 2 + 2 b a + 2 b a + 2 n n b a + 2 n n b a + ¥ ® n lim ¥ ® n lim î ¥ ® n lim ³ ¥ ® n lim £ î î î ¥ î k n x k n n x ¥ ® lim 0 x k n ® 0 x k n x k n x ® +¥ ; 2 p 2 p …

ma. (veyershtrassning ikkinchi teoremasi). agar f(x) funksiya [a;b] segmentda aniqlangan va uzluksiz bo`lsa, funksiya shu segmentda o`zining aniq quyi va aniq yuqori chegaralariga yerishadi. isbot. teoremaning xulosasini quyidagicha aytish mumkin, ya`ni [a;b] segmentda shunday x1 va x2 nuqtalar topiladiki, f(x1)= {f(x)}, f(x2)= {f(x)} bo`ladi (ya`ni f(x1) - f(x) funksiyaning [a;b] segmentdagi eng katta qiymati, f(x2) esa eng kichik qiymati). f(x) funksiya [a;b] da uzluksiz bo`lgani uchun veyershtrassning birinchi teoremasiga binoan f(x) [a;b] da chegaralangan, demak aniq yuqori va aniq quyi chegaralarga ega: {f(x)}=m , {f(x)}=m deylik. endi [a;b] da biror x1 nuqtasi uchun f(x1)=m bo`lishini ko`rsatamiz. teskarisini faraz qilaylik, ya`ni barcha x [a;b] larda f(x) 0 son topilib, ((x) ( bo`ladi. bundan f(x) x2 bo`lsin. u holda y=f(x) funksiya qat`iy o`suvchi bo`lganligi uchun f(x1)>f(x2), ya`ni y1>y2 bo`ladi. bu esa y1 0 uchun shunday >0 son topilib, |x- | 0 ga bog`liq bo`lib, nuqtaga bog`liq emas. ta`rif: agar har bir >0 …

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1576482411.doc 0 x 0 x e 0 x e £ 0 x £ 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x î 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 2 , b a a + b b a ; 2 + 2 b a + 2 b a + 2 n n b a + 2 n n b a + ¥ ® n lim ¥ ® n lim î ¥ ® n lim ³ ¥ ® n lim £ î î î ¥ î …

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