Buni amalga oshirish uchun biz qilishimiz kerak bo'lgan yagona narsa \(x\) ning har ikkala maxrajdan maxrajdagi eng katta quvvatini hisobga olishdir. va surat. Keyin biz qilishimiz kerak bo'lgan yagona narsa asosiy chegara xususiyatlaridan foydalanishdir 1-fakt limitni hisoblash uchun ushbu bo'limdan.

Maxraj uchun biz biroz ehtiyot bo'lishimiz kerak. Maxrajdagi \(x\) kuchi ildizdan tashqarida bo'lishi kerak, shuning uchun u suratdagi \(x\) ni bekor qilishi mumkin. \(x\) ning ildizdan tashqaridagi eng katta quvvati biz olishimiz mumkin (va ildizda hal qilishimiz mumkin bo'lgan narsani qoldirish) faqat \({x^2}\) bo'ladi. Biz buni ildizdan \({x^4}\) ko'paytuvchilarga ajratish orqali olamiz.

Shunday qilib, keling, bizni boshlash uchun ushbu jarayondagi dastlabki ikki qadamni qilaylik.

\[\begin{align*}\mathop {\lim }\limits_{x \to \, - \infty } \frac{{8 + x - 4{x^2}}}{{\sqrt {6 + {x^2} + 7{x^4}} }} & = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^2}\left( {\frac{8}{{{x^2}}} + \frac{1}{x} - 4} \right)}}{{\sqrt {{x^4}\left( {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} \right)} }}\\ & = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^2}\left( {\frac{8}{{{x^2}}} + \frac{1}{x} - 4} \right)}}{{\sqrt {{x^4}} \sqrt {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^2}\left( {\frac{8}{{{x^2}}} + \frac{1}{x} - 4} \right)}}{{\left| {{x^2}} \right|\sqrt {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} }}\end{align*}\]

Ushbu bo'limdagi muhokamadan shuni eslang.

\[\sqrt {{x^2}} = \left| x \right|\]

Shunday qilib, bu holda bizda bo'ladi.

\[\sqrt {{x^4}} = \left| {{x^2}} \right| = {x^2}\]

va biz mutlaq qiymat satrlaridan xalos bo'lishimiz mumkinligini unutmang, chunki biz \({x^2} \ge 0\) ekanligini bilamiz. Shunday qilib, keling, chegarani tugatamiz.

\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{8 + x - 4{x^2}}}{{\sqrt {6 + {x^2} + 7{x^4}} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^2}\left( {\frac{8}{{{x^2}}} + \frac{1}{x} - 4} \right)}}{{{x^2}\sqrt {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{\frac{8}{{{x^2}}} + \frac{1}{x} - 4}}{{\sqrt {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} }} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{ - 4}}{{\sqrt 7 }}}}\]

Oldingi ikkita muammodan farqli o'laroq, ulardagi ildizlar bilan bog'liq barcha matematik manipulyatsiyalar bu holatda haqiqiy chegaraga bog'liq emas edi, chunki biz \({x^2}\)ni ko'paytuvchilarga ajratish qildik, bu har doim ijobiy bo'ladi va shuning uchun bu ishni qayta bajarish uchun hech qanday sabab bo'lmaydi.

Mana bu chegara (ishlarning ko'p qismi bundan mustasno).

Ushbu qism uchun biz birinchi qismda mutlaq qiymat bilan ishlashgacha qilgan barcha matematik manipulyatsiyalar chegaraning o'ziga bog'liq emas edi va shuning uchun bu erda qayta ishlash kerak emas. Demak, bu qismga qadar bizda bor.

\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{8 + x - 4{x^2}}}{{\sqrt {6 + {x^2} + 7{x^4}} }} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{{x^2}\left( {\frac{8}{{{x^2}}} + \frac{1}{x} - 4} \right)}}{{{x^2}\sqrt {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} }} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{\frac{8}{{{x^2}}} + \frac{1}{x} - 4}}{{\sqrt {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} }} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{ - 4}}{{\sqrt 7 }}}}\]

Biz bilamizki, \(x \to - \infty \) uchun gorizontal asimptota bo'ladi, agar \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\) mavjud bo'lsa va cheklangan son bo'lsa. Xuddi shunday, agar \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\) mavjud bo'lsa va cheklangan son bo'lsa, biz \(x \to \infty \) uchun gorizontal asimptotaga ega bo'lamiz.

Shunday qilib, dastlabki ikki qismdan biz gorizontal asimptota olishimizni ko'rishimiz mumkin.

\[y = - \frac{4}{{\sqrt 7 }}\]

\(x \to - \infty \) va \(x \to \infty \) uchun.