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. \(x\) ning maxrajdagi kuchi ildizdan tashqarida bo'lishi kerak, shuning uchun u numeratordagi \(x\) ni bekor qilishi mumkin. \(x\) ning ildizdan tashqarida biz olishimiz mumkin bo'lgan (va ildizda hal qilishimiz mumkin bo'lgan narsani qoldiradigan) eng katta kuchi faqat \(x\) bo'ladi. Biz buni ildizdan \({x^2}\) ko'paytuvchilarga ajratish orqali olamiz.

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

\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{x + 8}}{{\sqrt {2{x^2} + 3} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{x\left( {1 + \frac{8}{x}} \right)}}{{\sqrt {{x^2}\left( {2 + \frac{3}{{{x^2}}}} \right)} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{x\left( {1 + \frac{8}{x}} \right)}}{{\sqrt {{x^2}} \sqrt {2 + \frac{3}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{x\left( {1 + \frac{8}{x}} \right)}}{{\left| x \right|\sqrt {2 + \frac{3}{{{x^2}}}} }}\]

Ushbu bo'limdagi muhokamadan shuni eslang.

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

va biz do bundan ehtiyot bo'lish kerak.

Endi biz \(x \to - \infty \) chegarasini ko'rib chiqayotganimiz uchun \(x < 0\) deb taxmin qilish mumkin. Shuning uchun, mutlaq qiymatning ta'rifidan biz olamiz.

\[\left| x \right| = - x\]

va chegara o'shanda.

\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{x + 8}}{{\sqrt {2{x^2} + 3} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{x\left( {1 + \frac{8}{x}} \right)}}{{ - x\sqrt {2 + \frac{3}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{1 + \frac{8}{x}}}{{ - \sqrt {2 + \frac{3}{{{x^2}}}} }} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{ - \sqrt 2 }}}}\]

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.

\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{x + 8}}{{\sqrt {2{x^2} + 3} }} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{x\left( {1 + \frac{8}{x}} \right)}}{{\left| x \right|\sqrt {2 + \frac{3}{{{x^2}}}} }}\]

Ushbu qismda biz \(x \to \infty \) chegarasini ko'rib chiqamiz va shuning uchun bu qismda \(x > 0\) deb taxmin qilish xavfsiz bo'ladi. Shuning uchun, mutlaq qiymatning ta'rifidan biz olamiz.

\[\left| x \right| = x\]

va chegara o'shanda.

\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{x + 8}}{{\sqrt {2{x^2} + 3} }} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{x\left( {1 + \frac{8}{x}} \right)}}{{x\sqrt {2 + \frac{3}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{1 + \frac{8}{x}}}{{\sqrt {2 + \frac{3}{{{x^2}}}} }} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{\sqrt 2 }}}}\]

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.

\[y = - \frac{1}{{\sqrt 2 }}\]

\(x \to - \infty \) uchun va bizda gorizontal asimptota bor.

\[y = \frac{1}{{\sqrt 2 }}\]

\(x \to \infty \) uchun.