2.6-bo'lim : Cheksiz limitlar
3. \(\displaystyle g\left( z \right) = \frac{{z + 3}}{{{{\left( {z + 1} \right)}^2}}}\) baholash uchun,
- \(\mathop {\lim }\limits_{z \to \, - {1^{\, - }}} g\left( z \right)\)
- \(\mathop {\lim }\limits_{z \to \, - {1^{\, + }}} g\left( z \right)\)
- \(\mathop {\lim }\limits_{z \to \, - 1} g\left( z \right)\)
a \(\mathop {\lim }\limits_{z \to \, - {1^{\, - }}} g\left( z \right)\) Yechim
Keling, \(z \to - {1^ - }\) uchun biz \(z < - 1\)ni bilishimizni tan olishdan boshlaylik.
Surat uchun biz chegarada 2 ni olishimizni ko'rishimiz mumkin.
Endi maxrajga g'amxo'rlik qilaylik. Cheklovda biz ega bo'lamiz,
\[z + 1 \to {0^ - }\]va \(z + 1\) ni kvadratga aylantirganimizda, biz chegarada,
\[{\left( {z + 1} \right)^2} \to {0^ + }\]Demak, chegarada surat musbat songa yaqinlashmoqda va maxraj borgan sari kichikroq ijobiy sondir. Bunda ko'rsatkich ortib borayotgan katta musbat son bo'lishi kerak yoki,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{z \to \, - {1^{\, - }}} \frac{{z + 3}}{{{{\left( {z + 1} \right)}^2}}} = \infty }}\]E'tibor bering, bu \(z = - 1\) da vertikal asimptota mavjudligini ham anglatadi.
b \(\mathop {\lim }\limits_{z \to \, - {1^{\, + }}} g\left( z \right)\) Yechim
Keling, \(z \to - {1^ + }\) uchun biz \(z > - 1\)ni bilishimizni tan olishdan boshlaylik.
Surat uchun biz chegarada 2 ni olishimizni ko'rishimiz mumkin.
Endi maxrajga g'amxo'rlik qilaylik. Cheklovda biz ega bo'lamiz,
\[z + 1 \to {0^ + }\]va \(z + 1\) ni kvadratga aylantirgandan so'ng, biz chegarada bo'lishini ko'ramiz,
\[{\left( {z + 1} \right)^2} \to {0^ + }\]Demak, chegarada surat musbat songa yaqinlashmoqda va maxraj borgan sari kichikroq ijobiy sondir. Bunda ko'rsatkich ortib borayotgan katta musbat son bo'lishi kerak yoki,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{z \to \, - {1^{\, + }}} \frac{{z + 3}}{{{{\left( {z + 1} \right)}^2}}} = \infty }}\]E'tibor bering, bu \(z = - 1\) da vertikal asimptota mavjudligini ham anglatadi, biz buni birinchi qismdan allaqachon bildik.
c \(\mathop {\lim }\limits_{z \to \, - 1} g\left( z \right)\) Yechim
Bu holda biz dastlabki ikki qismdan shuni ko'rishimiz mumkinki,
\[\mathop {\lim }\limits_{z \to \, - {1^{\, - }}} g\left( z \right) = \mathop {\lim }\limits_{z \to \, - {1^{\, + }}} g\left( z \right) = \infty \]va shuning uchun bizning asosiy chegara xususiyatlaridan shuni ko'rishimiz mumkinki,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{z \to \, - 1} g\left( z \right) = \infty }}\]To'liqlik va ushbu muammoning javoblarini tekshirish uchun bu erda funktsiyaning qisqacha eskizi keltirilgan.
