Isbot. Har ikkala funksiyani x=a nuqtada f(a)=0, g(a)=0 deb aniqlasak, natijada ikkinchi shartga ko‘ra f(x)=0=f(a), g(x)=0=g(a) tengliklar o‘rinli bo‘lib, f(x) va g(x) funksiyalar x=a nuqtada uzluksiz bo‘ladi.
Avval x>a holni qaraymiz. Berilgan f(x) va g(x) funksiyalar [a;x], bu erda x, kesmada Koshi teoremasining shartlarini qanoatlantiradi. Shuning uchun a bilan x orasida shunday c nuqta topiladiki, ushbu tenglik o‘rinli bo‘ladi. f(a)=g(a)=0 ekanligini e’tiborga olsak, so‘ngi tenglikdan
(2)
bo‘lishi kelib chiqadi. Ravshanki,abo‘lganligi sababli, xa bo‘lganda ca bo‘ladi. Teoremaning 3-sharti va (2) tenglikdan = =A kelib chiqadi.
Shunga o‘xshash, x holni ham qaraladi. Teorema isbot bo‘ldi.
Misol. Ushbu limitni xisoblang.
Yechish.Bu holda bo‘lib, ular uchun 1- teoremaning barcha shartlari bajariladi.
Haqiqatan ham,
1) , ;
2) ;
3) bo‘ladi.
Demak, 1-teoremaga binoan .
1-eslatma. Shuni ta’kidlash kerakki, berilgan funksiyalar nisbatining limiti 3) shart bajarilmasa ham mavjud bo‘lishi mumkin, ya’ni 3) shart yyetarli bo‘lib, zaruriy emas.
Masalan, funksiyalar (0;1] da 1), 2) shartlarni qanoatlantiradi va , lekin
mavjud emas, chunki n da
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.
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