2.10-bo'lim : Limitning ta'rifi
4. Quyidagi chegarani isbotlash uchun chegara ta’rifidan foydalaning.
\[\mathop {\lim }\limits_{x \to - 3} \left( {{x^2} + 4x + 1} \right) = - 2\]Yechim
Birinchidan, nima ko'rsatishimiz kerakligini yozamiz.
\(\varepsilon > 0\) istalgan raqam bo'lsin. Biz \(\delta > 0\) raqamini topishimiz kerak, shunda
\[\left| {{x^2} + 4x + 1 - \left( { - 2} \right)} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - \left( { - 3} \right)} \right| < \delta \]Buni biroz soddalashtirsak,
\[\left| {{x^2} + 4x + 3} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x + 3} \right| < \delta \]2-qadam
Birinchi tengsizlikni biroz soddalashtirishdan boshlaylik.
\[\left| {{x^2} + 4x + 3} \right| = \left| {\left( {x + 1} \right)\left( {x + 3} \right)} \right| = \left| {x + 1} \right|\left| {x + 3} \right| < \varepsilon \]Bizda biz ko'rishni kutayotgan \(\left| {x + 3} \right|\) bor, lekin bizda \(\left| {x + 1} \right|\) ham bor, u bilan shug'ullanishimiz kerak.
3-qadam
\(\left| {x + 1} \right|\) bilan ishlash uchun, avvalo, buni taxmin qilaylik
\[\left| {x + 3} \right| < 1\]Ushbu bo'lim uchun eslatmalarda shunga o'xshash misolda ta'kidlaganimizdek, bu qonuniy taxmin, chunki chegara \(x \to - 3\) va shuning uchun \(x\) -3 ga juda yaqinlashadi. Shuning uchun, \(x\) -3 ga etarlicha yaqin bo'lsa, bizda \(\left| {x + 3} \right| < 1\) bo'ladi.
Ushbu taxmindan boshlab, biz shuni tushunamiz,
\[ - 1 < x + 3 < 1\hspace{0.25in} \to \hspace{0.25in}\,\, - 4 < x < - 2\] Agar biz ushbu tengsizlikning barcha qismlariga 1 ni qo'shsak, \[ - 3 < x + 1 < - 1\]-1 < 0 ekanligini e'tiborga olsak, biz \(x + 1 < 0\) va shuning uchun \(\left| {x + 3} \right| < 1\) taqdim etilganda bizda \(\left| {x + 1} \right| = - \left( {x + 1} \right)\) bo'lishini ham bilib olamiz. Bundan tashqari, yuqoridagi tengsizlikdan shuni ko'ramizki,
\[1 < - \left( {x + 1} \right) < 3\]Bularning barchasi shuni anglatadiki, \(\left| {x + 3} \right| < 1\) sharti bilan bizda ham quyidagilar bo'ladi:
\[\left| {x + 1} \right| = - \left( {x + 1} \right) < 3\hspace{0.5in} \to \hspace{0.5in}\left| {x + 1} \right| < 3\]Bu, o'z navbatida, bizda borligini anglatadi,
\[\left| {x + 1} \right|\left| {x + 3} \right| < 3\left| {x + 3} \right|\hspace{0.25in}\,\,\,\,\,\,{\mbox{because }}\left| {x + 1} \right| < 3\]Shuning uchun, agar biz biron sababga ko'ra \(3\left| {x + 3} \right| < \varepsilon \) ni xohlaymiz deb taxmin qilsak, bu bizga shuni aytadi:
\[\left| {x + 3} \right| < \frac{\varepsilon }{3}\]4-qadam
Mayli, bu kabi ko'rinmasa ham, bizda \(\delta \) uchun tanlov qilish uchun yetarlicha narsa bor.
Har qanday \(\varepsilon > 0\) raqamini hisobga olsak, keling, tanlaymiz
\[\delta = \min \left\{ {1,\frac{\varepsilon }{3}} \right\}\]Shunga qaramay, bu \(\delta \) ikkita qiymatdan kichikroq bo'lishini anglatadi, bu esa o'z navbatida shuni anglatadiki,
\[\delta \le 1\hspace{0.5in}{\mbox{AND}}\hspace{0.5in}\delta \le \frac{\varepsilon }{3}\]Endi \(0 < \left| {x + 3} \right| < \delta = \min \left\{ {1,\frac{\varepsilon }{3}} \right\}\) deb faraz qiling.
5-qadam
Shunday qilib, keling, bu ish yoki yo'qligini bilib olaylik.
\(0 < \left| {x + 3} \right| < \delta = \min \left\{ {1,\frac{\varepsilon }{3}} \right\}\) taxminini hisobga olsak, biz ikkita narsani bilamiz. Birinchidan, biz \(\left| {x + 3} \right| < \frac{\varepsilon }{3}\) ekanligini bilamiz. Ikkinchidan, biz \(\left| {x + 3} \right| < 1\) ni ham bilamiz, bu o'z navbatida 3-bosqichda ko'rganimizdek \(\left| {x + 1} \right| < 3\) ni anglatadi.
Endi quyidagilarni bajaramiz,
\[\begin{align*}\left| {{x^2} + 4x + 3} \right| & = \left| {x + 1} \right|\left| {x + 3} \right| & & \hspace{0.25in}{\mbox{factoring}}\\ & < 3\left| {x + 3} \right| & & \hspace{0.25in}{\mbox{because we know }}\left| {x + 1} \right| < 3\\ & < 3\left( {\frac{\varepsilon }{3}} \right) & & \hspace{0.25in}\,{\mbox{because we know }}\left| {x + 3} \right| < \frac{\varepsilon }{3}\\ & = \varepsilon & & \end{align*}\]Shunday qilib, biz buni ko'rsatdik,
\[\left| {{x^2} + 4x + 3} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x + 3} \right| < \min \left\{ {1,\frac{\varepsilon }{3}} \right\}\]va shuning uchun chegaraning ta'rifi bilan biz buni isbotladik,
\[\mathop {\lim }\limits_{x \to - 3} \left( {{x^2} + 4x + 1} \right) = - 2\]