Asosiy tarkibga o'tish

2.6-bo'lim : Cheksiz limitlar

2. \(\displaystyle h\left( t \right) = \frac{{2t}}{{6 + t}}\) baholash uchun,

  1. \(\mathop {\lim }\limits_{t \to \, - {6^{\, - }}} h\left( t \right)\)
  2. \(\mathop {\lim }\limits_{t \to \, - {6^{\, + }}} h\left( t \right)\)
  3. \(\mathop {\lim }\limits_{t \to \, - 6} h\left( t \right)\)
a \(\mathop {\lim }\limits_{t \to \, - {6^{\, - }}} h\left( t \right)\) Yechim

Keling, \(t \to - {6^ - }\) uchun biz \(t < - 6\)ni bilishimizni tan olishdan boshlaylik.

Surat uchun biz chegarada -12 ni olishimizni ko'rishimiz mumkin.

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

\[6 + t \to 0\]

lekin biz aslida biroz uzoqroqqa borishimiz mumkin. Biz \(t < - 6\) ekanligini bilganimiz uchun biz buni ham bilamiz,

\[6 + t < 0\]

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

\[6 + t \to {0^ - }\]

Demak, chegarada surat manfiy songa yaqinlashmoqda va maxraj borgan sari kichikroq manfiy sondir. Bunda ko'rsatkich ortib borayotgan katta musbat son bo'lishi kerak yoki,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{t \to \, - {6^{\, - }}} \frac{{2t}}{{6 + t}} = \infty }}\]

E'tibor bering, bu \(t = - 6\) da vertikal asimptota mavjudligini ham anglatadi.


b \(\mathop {\lim }\limits_{t \to \, - {6^{\, + }}} h\left( t \right)\) Yechim

Keling, \(t \to - {6^ + }\) uchun biz \(t > - 6\)ni bilishimizni tan olishdan boshlaylik.

Surat uchun biz chegarada -12 ni olishimizni ko'rishimiz mumkin.

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

\[6 + t \to 0\]

lekin biz aslida biroz uzoqroqqa borishimiz mumkin. Biz \(t > - 6\) ekanligini bilganimiz uchun biz buni ham bilamiz,

\[6 + t > 0\]

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

\[6 + t \to {0^ + }\]

Demak, chegarada surat manfiy songa yaqinlashmoqda va maxraj borgan sari kichikroq ijobiy sondir. Bo'lim katta manfiy son bo'lishi kerak yoki,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{t \to \, - {6^{\, + }}} \frac{{2t}}{{6 + t}} = - \infty }}\]

E'tibor bering, bu \(t = - 6\) da vertikal asimptota mavjudligini ham anglatadi, biz uni birinchi qismdan bildik.


c \(\mathop {\lim }\limits_{t \to \, - 6} h\left( t \right)\) Yechim

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

\[\mathop {\lim }\limits_{t \to \, - {6^{\, - }}} h\left( t \right) \ne \mathop {\lim }\limits_{t \to \, - {6^{\, + }}} h\left( t \right)\]

Shunday qilib, bizning asosiy chegara xususiyatlaridan biz \(\mathop {\lim }\limits_{t \to \, - 6} h\left( t \right)\) ekanligini ko'rishimiz mumkin mavjud emas.

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

The graph of \(h\left( t \right) = \frac{{2t}}{{6 + t}}\).  There is a vertical asymptote at \(t = - 6\).  To the left of it the graph rises steeply to positive infinity as \(t\) approaches -6, and to the right of it the graph comes up from negative infinity, crosses the t-axis at the origin and flattens out towards the horizontal asymptote \(h = 2\).