Buni amalga oshirish uchun biz qilishimiz kerak bo'lgan yagona narsa \(x\) ning eng katta kuchini butun ko'phaddan ajratib ko'rsatish va keyin asosiy chegara xususiyatlaridan foydalanish. 1-fakt limitni hisoblash uchun ushbu bo'limdan.

Shuni ham yodda tutingki, biz ildizni kasr ko'rsatkichiga aylantiramiz, bu bilan ishlash osonroq bo'ladi. Shuni ham yodda tutingki, bu chegara mukammal qabul qilinadigan chegaradir, chunki ildiz kub ildizi va biz mumkin manfiy sonlarning kub ildizlarini oling! Ildizdagi indeks juft son bo'lgandagina muammolarga duch kelgan bo'lardik.

\[\begin{align*}\mathop {\lim }\limits_{t \to \, - \infty } \left( {{t^{\frac{1}{3}}} + 12t - 2{t^2}} \right) & = \mathop {\lim }\limits_{t \to \, - \infty } \left[ {{t^2}\left( {\frac{1}{{{t^{\frac{5}{3}}}}} + \frac{{12}}{t} - 2} \right)} \right]\\ & = \left( {\mathop {\lim }\limits_{t \to \, - \infty } {t^2}} \right)\left[ {\mathop {\lim }\limits_{t \to \, - \infty } \left( {\frac{1}{{{t^{\frac{5}{3}}}}} + \frac{{12}}{t} - 2} \right)} \right] = \left( \infty \right)\left( { - 2} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \infty }}\end{align*}\]

Ushbu qism uchun biz birinchi qismda qilgan barcha matematik manipulyatsiyalar chegaraning o'ziga bog'liq emas edi va shuning uchun bu erda takrorlash shart emas. Biz muammoni chegaralarni olishimizdan oldin hal qilishimiz va keyin davom etishimiz mumkin.

\[\mathop {\lim }\limits_{t \to \,\infty } \left( {{t^{\frac{1}{3}}} + 12t - 2{t^2}} \right) = \left( {\mathop {\lim }\limits_{t \to \,\infty } {t^2}} \right)\left[ {\mathop {\lim }\limits_{t \to \,\infty } \left( {\frac{1}{{{t^{\frac{5}{3}}}}} + \frac{{12}}{t} - 2} \right)} \right] = \left( \infty \right)\left( { - 2} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \infty }}\]