Asosiy tarkibga o'tish

2.6-bo'lim : Cheksiz limitlar

1. \(\displaystyle f\left( x \right) = \frac{9}{{{{\left( {x - 3} \right)}^5}}}\) baholash uchun,

  1. \(\mathop {\lim }\limits_{x \to {3^{\, - }}} f\left( x \right)\)
  2. \(\mathop {\lim }\limits_{x \to {3^{\, + }}} f\left( x \right)\)
  3. \(\mathop {\lim }\limits_{x \to 3} f\left( x \right)\)
a \(\mathop {\lim }\limits_{x \to {3^{\, - }}} f\left( x \right)\) Yechim

Keling, \(x \to {3^ - }\) uchun biz \(x < 3\)ni bilishimizni tan olishdan boshlaylik.

Surat uchun biz chegarada faqat 9 bo'lishini ko'rishimiz mumkin.

Maxraj biroz ko'proq ish talab qiladi. Shubhasiz, chegarada bizda,

\[x - 3 \to 0\]

lekin biz aslida biroz uzoqroqqa borishimiz mumkin. Biz \(x < 3\) ekanligini bilganimiz uchun biz buni ham bilamiz,

\[x - 3 < 0\]

Keyinchalik ixchamroq qilib aytishimiz mumkinki, biz chegarada bo'lamiz,

\[x - 3 \to {0^ - }\]

Buni beshinchi darajaga ko'tarish bu xatti-harakatni o'zgartirmaydi va shuning uchun chegarada maxraj bo'ladi:

\[{\left( {x - 3} \right)^5} \to {0^ - }\]

Endi biz funktsiya chegarasini bajarishimiz mumkin. Chegarada surat sobit musbat doimiy, maxraj esa tobora kichikroq manfiy sondir. Limitda bo'linuvchi katta manfiy son bo'lishi kerak yoki,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to {3^{\, - }}} \frac{9}{{{{\left( {x - 3} \right)}^5}}} = - \infty }}\]

E'tibor bering, bu \(x = 3\) da vertikal asimptota mavjudligini ham anglatadi.


b \(\mathop {\lim }\limits_{x \to {3^{\, + }}} f\left( x \right)\) Yechim

Keling, \(x \to {3^ + }\) uchun biz \(x > 3\)ni bilishimizni tan olishdan boshlaylik.

Birinchi qismda bo'lgani kabi, surat, chegarada u faqat 9 bo'ladi.

Maxraj ham birinchi qismga o'xshab ishlaydi. Bizda chegarada,

\[x - 3 \to 0\]

va biz \(x > 3\) ekanligini bilganimiz uchun biz buni ham bilamiz,

\[x - 3 > 0\]

Keyinchalik ixchamroq qilib aytishimiz mumkinki, biz chegarada bo'lamiz,

\[x - 3 \to {0^ + }\]

Buni beshinchi darajaga ko'tarish bu xatti-harakatni o'zgartirmaydi va shuning uchun chegarada maxraj bo'ladi:

\[{\left( {x - 3} \right)^5} \to {0^ + }\]

Endi biz funktsiya chegarasini bajarishimiz mumkin. Chegarada surat sobit musbat doimiy, maxraj esa tobora kichikroq musbat sondir. Limitda bo'linuvchi katta musbat son bo'lishi kerak yoki,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to {3^{\, + }}} \frac{9}{{{{\left( {x - 3} \right)}^5}}} = \infty }}\]

E'tibor bering, bu \(x = 3\) da vertikal asimptota mavjudligini anglatadi, biz buni birinchi qismdan allaqachon bildik.


c \(\mathop {\lim }\limits_{x \to 3} f\left( x \right)\) Yechim

Bu holda biz dastlabki ikki qismdan shuni ko'rishimiz mumkinki,

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

Shunday qilib, bizning asosiy chegara xususiyatlaridan biz \(\mathop {\lim }\limits_{x \to 3} f\left( x \right)\) ekanligini ko'rishimiz mumkin mavjud emas.

To'liqlik va ushbu muammoning javoblarini tekshirish uchun bu erda funktsiyaning qisqacha eskizi keltirilgan.

The graph of \(f\left( x \right) = \frac{9}{{{{\left( {x - 3} \right)}^5}}}\).  There is a vertical asymptote at \(x = 3\).  To the left of it the graph runs along just below the x-axis and then drops away steeply to negative infinity as \(x\) approaches 3, and to the right of it the graph comes down from positive infinity and then flattens out just above the x-axis.