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2.5-bo'lim : Hisoblash chegaralari

14. \(x \le 3\) uchun \(x^{3}-6x^{2}+12x-3 \le f\left( x \right) \le x^{2}-4x+9\) \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) qiymatini aniqlashini hisobga olsak.

Yechim

Birinchidan, biz \(f\left( x \right)\) har doim \( x\le 3\) uchun boshqa ikkita funksiya orasida ekanligini va \(x = 2\) bu oraliqda ekanligini hisobga olsak, bu muammo siqish teoremasi uchun o'rnatiladi. Endi biz qilishimiz kerak bo'lgan narsa bu ikkita "tashqi" funktsiyaning \(x = 2\) da bir xil chegaraga ega ekanligini tekshirish va agar ular bo'lsa, javob olish uchun siqish teoremasidan foydalanishimiz mumkin.

\[\mathop {\lim }\limits_{x \to 2} \left(x^{3}-6x^{2}+12x-3\right) = 5\hspace{0.75in}\mathop {\lim }\limits_{x \to 2} \left( x^{2}-4x+9 \right) = 5\]

Demak, bizda bor,

\[\mathop {\lim }\limits_{x \to 2} \left(x^{3}-6x^{2}+12x-3\right) = \mathop {\lim }\limits_{x \to 2} \left( x^{2}-4x+9\right) = 5\]

va shuning uchun siqish teoremasi bo'yicha biz ham ega bo'lishimiz kerak,

\[\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 5\]