sonlar ketma – ketligi

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1662970009.doc a x n n = ¥ ® lim ( a x yoki n = lim ) a x yoki n ® a x n n = ¥ ® lim ¥ = ¥ ® n n x lim ¥ = ¥ ® n n 2 lim 0 lim = ¥ ® n n x a x n n - = a n x 2 1 3 1 4 1 n 1 n n a x a + = n a 0 > " e e + = n n x n n n n 1 1 1 + = + n 1 2 1 1 e ) ( e n ) ( e n n > e e n ) ( e n 2 1 1 - e ÷ ø ö ç è æ - e = 2 1 1 e n 0 > e e e e => + …

4+() intervaldan iborat; ( ga istalgan musbat qiymatlarni berib, koordinatasi 4 ga teng nuqtaning hamma atroflarini hosil qilamiz. tanlangan ( uchun 4-( 0 son olinganda ham shunday natural n0 son topilib, barcha n > n0 uchun a-( 0 da ( x – c( =( c – c( = 0 n0 uchun ( xn( >m tengsizlik o’rinli bo’lsa, {xn} ketma – ketlikning limiti deb qaraladi va yoki kabi belgilanadi. agar har qanday m > 0 son berilganda ham shunday n0(n son topilsaki, barcha n > n0 uchun xn > m(xn n bo’lganda bajariladi, bu esa . b) endi absalyut miqdor farqini topamiz: . berilgan n ni shunday tanlaymizki tengsizlik bajarilsin. bu tengsizlikni yechib: . deb, da . agar bo’lsa, , bundan ketma – ketlikning 6-hadidan boshlab barcha hadlari intervalda yotadi. 2 – m i s o l. ushbu 0,1; 0,11; 0,111; … ketma – ketlikning limiti ga teng. i s b …

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1662970009.doc a x n n = ¥ ® lim ( a x yoki n = lim ) a x yoki n ® a x n n = ¥ ® lim ¥ = ¥ ® n n x lim ¥ = ¥ ® n n 2 lim 0 lim = ¥ ® n n x a x n n - = a n x 2 1 3 1 4 1 n 1 n n a x a + = n a 0 > " e e + = n n x n n n n 1 1 1 + = + n 1 2 1 1 e ) ( e n ) ( e n n > e e n ) …

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