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.

Bu holda \(x\) ning maxrajdagi eng katta kuchi shunchaki \(x\) dir va shuning uchun biz \(x\) ni maxrajdan ham, suratdan ham ajratib olishimiz kerak bo'ladi. Shuni ham esda tutingki, bu biz suratdagi ildizdan \({x^2}\) ni koeffitsient qilishimiz kerakligini anglatadi, shunda biz tugatgandan so'ng numeratorda \(x\)ga ega bo'lamiz.

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

\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{\sqrt {7 + 9{x^2}} }}{{1 - 2x}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{\sqrt {{x^2}\left( {\frac{7}{{{x^2}}} + 9} \right)} }}{{x\left( {\frac{1}{x} - 2} \right)}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{\sqrt {{x^2}} \sqrt {\frac{7}{{{x^2}}} + 9} }}{{x\left( {\frac{1}{x} - 2} \right)}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{\left| x \right|\sqrt {\frac{7}{{{x^2}}} + 9} }}{{x\left( {\frac{1}{x} - 2} \right)}}\]

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.

\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{\sqrt {7 + 9{x^2}} }}{{1 - 2x}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{ - x\sqrt {\frac{7}{{{x^2}}} + 9} }}{{x\left( {\frac{1}{x} - 2} \right)}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{ - \sqrt {\frac{7}{{{x^2}}} + 9} }}{{\frac{1}{x} - 2}} = \frac{{ - \sqrt 9 }}{{ - 2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{3}{2}}}\]

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.

\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{\sqrt {7 + 9{x^2}} }}{{1 - 2x}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{\left| x \right|\sqrt {\frac{7}{{{x^2}}} + 9} }}{{x\left( {\frac{1}{x} - 2} \right)}}\]

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.

\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{\sqrt {7 + 9{x^2}} }}{{1 - 2x}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{x\sqrt {\frac{7}{{{x^2}}} + 9} }}{{x\left( {\frac{1}{x} - 2} \right)}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{\sqrt {\frac{7}{{{x^2}}} + 9} }}{{\frac{1}{x} - 2}} = \frac{{\sqrt 9 }}{{ - 2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{3}{2}}}\]

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.

\[y = \frac{3}{2}\]

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

\[y = - \frac{3}{2}\]

\(x \to \infty \) uchun.