Ushbu qism uchun biz \( - 6 < 1\) ekanligini bilamiz va shuning uchun qiymatlar bo'ladi x -6 ning ikkala tomonida \(x < 1\) oralig'ida va shuning uchun biz chegarada \(x < 1\) bo'ladi deb taxmin qilishimiz mumkin. Bu bizga ushbu diapazondagi funktsiya qismini ishlatishga imkon beradi va keyin chegarani hisoblash uchun standart chegara usullaridan foydalaning.

\[\mathop {\lim }\limits_{x \to \, - 6} f\left( x \right) = \mathop {\lim }\limits_{x \to \, - 6} \left( {7 - 4x} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{31}}\]

Bu qism oldingi qismdan farq qiladi. Biz \(x = 1\)-dagi chegarani ko'rib chiqamiz va bu qismlarga bo'lingan funktsiyalardagi "kesish" nuqtasidir. Bo'limdagi muhokamadan eslang, bu biz ikkita bir tomonlama chegaralarni ko'rib chiqishimiz kerakligini anglatadi.

\[\mathop {\lim }\limits_{x \to {1^{\, - }}} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^{\, - }}} \left( {7 - 4x} \right) = \underline 3 \hspace{0.25in}\hspace{0.25in}{\mbox{because }}x \to {1^ - }{\mbox{ implies that }}x < 1\] \[\mathop {\lim }\limits_{x \to {1^{\, + }}} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^{\, + }}} \left( {{x^2} + 2} \right) = \underline 3 \hspace{0.25in}\hspace{0.25in}{\mbox{because }}x \to {1^ + }{\mbox{ implies that }}x > 1\]

Shunday qilib, bu holatda biz buni ko'rishimiz mumkin,

\[\mathop {\lim }\limits_{x \to {1^{\, - }}} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^{\, + }}} f\left( x \right) = 3\]

va shuning uchun biz umumiy chegara mavjudligini bilamiz va,

\[\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \require{bbox} \bbox[2pt,border:1px solid black]{3}\]