Lopital teoremasi

Lopital teoremasi (shuningdek, Bernulli — Lopital qoidasi) — bu $0/0$ va $\mathrm{\infty }/\mathrm{\infty }$ shaklining noaniqliklarini ochib beradigan funksiyalar chegaralarini topish usuli. Usulni asoslovchi teorema maʼlum sharoitlarda funksiyalar nisbati chegarasi ularning hosilalari nisbati chegarasiga teng ekanligini tasdiqlaydi.

Aniq ifodalanishi

Lopital teoremasi:

Agar: $f(x),g(x)$ $a$ nuqtaning $U$ teshilgan atrofida differensiallanadigan haqiqatan qiymatli funksiyalardir, buyerda $a$ haqiqiy son yoki $+\mathrm{\infty },-\mathrm{\infty },\mathrm{\infty }$ belgilaridan biri, buning ustiga

1. $\underset{x\to a}{\lim }f(x)=\underset{x\to a}{\lim }g(x)=0$ yoki $\mathrm{\infty }$;
2. $g^{\prime }(x)\ne 0$, $U$ ichida;
3. $\lim {\mathrm{\backslash limits}}_{x\to a}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ mavjud;

bunday holda $\lim {\mathrm{\backslash limits}}_{x\to a}\frac{f(x)}{g(x)}=\lim {\mathrm{\backslash limits}}_{x\to a}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ mavjud.

Chegaralar ham bir tomonlama bo‘lishi mumkin.

Tarix

Ushbu turdagi noaniqlikni hal qilish usuli Giyom Lopital tomonidan 1696 yilda „Analyse des Infiniment Petits“ darsligida nashr etilgan. Usul Lopitalga uning kashfiyotchisi Iogan Bernulli tomonidan maktubda xabar qilingan.

Misollar

• $\underset{x\to 0}{\lim }\frac{x^2+5x}{3x}=\underset{x\to 0}{\lim }\frac{2x+5}{3}=\frac{5}{3}$

• $\underset{x\to \mathrm{\infty }}{\lim }\frac{x^3+4x^2+7x+9}{x^3+3x^2}=\underset{x\to \mathrm{\infty }}{\lim }\frac{3x^2+8x+7}{3x^2+6x}=\underset{x\to \mathrm{\infty }}{\lim }\frac{6x+8}{6x+6}=\frac{6}{6}=1$
  Bu yerda Lopital qoidasini 3 marta qo‘llash mumkin, ammo boshqacha qilish ham mumkin. Surat va maxrajni eng katta darajasida $x$ ga bo‘lish kerak (bizning holatlarimizda $x^3$). Bu misol natijasi quyidagicha bo‘ladi:

  $\underset{x\to \mathrm{\infty }}{\lim }\frac{1+4/x+7/x^2+9/x^3}{1+3/x}=\frac{1}{1}=1$

• $\underset{x\to +\mathrm{\infty }}{\lim }\frac{e^x}{x^a}=\underset{x\to +\mathrm{\infty }}{\lim }\frac{e^x}{a\cdot x^{a-1}}=\dots =\underset{x\to +\mathrm{\infty }}{\lim }\frac{e^x}{a!}=+\mathrm{\infty }$ — $a$ marta qoidani qo‘llash;
• $\underset{x\to +\mathrm{\infty }}{\lim }\frac{x^a}{\ln x}=\underset{x\to +\mathrm{\infty }}{\lim }\frac{a{x}^{a-1}}{\frac{1}{x}}=a\cdot \underset{x\to +\mathrm{\infty }}{\lim }{x}^a=+\mathrm{\infty }$ $a>0$ da;
• $\underset{x\to +\mathrm{\infty }}{\lim }\frac{\underset{x}{\overset{+\mathrm{\infty }}{\int }}{e}^{-t^2}dt}{x^{-1}{e}^{-x^2}}=\underset{x\to +\mathrm{\infty }}{\lim }\frac{-e^{-x^2}}{-x^{-2}{e}^{-x^2}(1+2x^2)}=\underset{x\to +\mathrm{\infty }}{\lim }\frac{x^2}{1+2x^2}=\frac{1}{2}$ .

Natija

Lopital qoidasining oddiy, ammo foydali natijasi, funksiyalarning differentsialligi mezoni quyidagicha:

$a$ nuqtaning teshilgan atrofida $f(x)$ funksiyasi differensiallanuvchi bo‘lsin va aynan shu nuqtada u uzluksiz va hosilaviy $\underset{x\to a}{\lim }{f}^{\prime }(x)=A$ chegaraga ega. Bunda $f(x)$ funksiyasi $a$ nuqtaning o‘zida ham farqlanadi va $f^{\prime }(a)=A$ (yaʼni, $f^{\prime }(x)$ hosilasi $a$ nuqtada uzluksiz).

Buni isbotlash uchun Lopital qoidasini $\frac{f(x)-f(a)}{x-a}$ nisbatiga qo‘llash kifoya.