Bu juda oddiy dalil. Belgilanishni biroz aniqroq qilish uchun keling, \(f\left( x \right) = c\) funktsiyasini aniqlaymiz, keyin bizdan bu \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = c\) ekanligini isbotlashimiz so'raladi. Shunday qilib, keling, buni qilaylik.

\(\varepsilon > 0\) bo'lsin va biz \(\delta > 0\) ni topishimiz mumkinligini ko'rsatishimiz kerak, shunda

\[\left| {f\left( x \right) - c} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta \]

Chap tengsizlik har qanday \(x\) uchun ahamiyatsiz, ammo biz \(f\left( x \right) = c\) ni aniqlaganimiz uchun. Shunday qilib, siz xohlagan raqam bo'lish uchun \(\delta > 0\) ni tanlang (odatda bu dalillar bilan buni qilolmaysiz). Keyin,

\[\left| {f\left( x \right) - c} \right| = \left| {c - c} \right| = 0 < \varepsilon \]

Ushbu qismni isbotlashning bir necha yo'li mavjud. Qabul qilsangiz 3 Va 7 keyin siz qilishingiz kerak bo'lgan yagona narsa \(g\left( x \right) = c\) ga ruxsat berish va keyin bu to'g'ridan-to'g'ri natijadir 3 va 7. Biroq, biz yanada qattiqroq matematik dalil qilishni xohlaymiz. Demak, bu dalil.

Birinchidan, agar \(c = 0\), keyin \(cf\left( x \right) = 0\) va hokazo.

\[\mathop {\lim }\limits_{x \to a} \left[ {0f\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} 0 = 0 = 0f\left( x \right)\]

Limitni baholash alohida holatdir 7 (\(c = 0\) bilan) biz buni isbotladik, shuning uchun biz bilamiz 1 \(c = 0\) uchun to'g'ri va shuning uchun biz ushbu dalilning qolgan qismi uchun \(c \ne 0\) deb taxmin qilishimiz mumkin.

\(\varepsilon > 0\) bo'lsin, chunki \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = K\) chegara ta'rifi bo'yicha \({\delta _{\,1}} > 0\) mavjud, shundayki,

\[\left| {f\left( x \right) - K} \right| < \frac{\varepsilon }{{\left| c \right|}}\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,1}}\]

Endi \(\delta = {\delta _{\,1}}\) ni tanlang va biz buni ko'rsatishimiz kerak

\[\left| {cf\left( x \right) - cK} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta \]

va biz tugatamiz. Shunday qilib, \(0 < \left| {x - a} \right| < \delta \) deb faraz qiling va keyin,

\[\left| {cf\left( x \right) - cK} \right| = \left| c \right|\left| {f\left( x \right) - K} \right| < \left| c \right|\frac{\varepsilon }{{\left| c \right|}} = \varepsilon \]

Biz bu dalilni ikki qismda qilamiz. Avval \(\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) + g\left( x \right)} \right] = K + L\) ni isbotlaymiz.

\(\varepsilon > 0\) bo'lsin, chunki \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = K\) va \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = L\) mavjud \({\delta _{\,1}} > 0\) va \({\delta _2} > 0\) shundayki,

\[\begin{align*}\left| {f\left( x \right) - K} \right| & < \frac{\varepsilon }{2}\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta {\,_1}\\ \left| {g\left( x \right) - L} \right|& < \frac{\varepsilon }{2}\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,2}}\end{align*}\]

Endi \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) ni tanlang. Keyin buni ko'rsatishimiz kerak

\[\left| {f\left( x \right) + g\left( x \right) - \left( {K + L} \right)} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta \]

\(0 < \left| {x - a} \right| < \delta \) deb faraz qiling. Keyin bizda,

\[\begin{align*}\left| {f\left( x \right) + g\left( x \right) - \left( {K + L} \right)} \right| & = \left| {\left( {f\left( x \right) - K} \right) + \left( {g\left( x \right) - L} \right)} \right|\\ & \le \left| {f\left( x \right) - K} \right| + \left| {g\left( x \right) - L} \right|\hspace{0.25in}{\mbox{by the triangle inequality}}\\ & < \frac{\varepsilon }{2} + \frac{\varepsilon }{2}\\ & = \varepsilon \end{align*}\]

Uchinchi bosqichda biz \(\delta \) ni tanlagan holda bizda \(0 < \left| {x - a} \right| < {\delta _{\,1}}\) va \(0 < \left| {x - a} \right| < {\delta _{\,2}}\) ham mavjudligidan foydalandik va shuning uchun biz isbotlashda dastlabki bayonotlardan foydalanishimiz mumkin.

Keyinchalik, biz \(\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - g\left( x \right)} \right] = K - L\) ni isbotlashimiz kerak. Biz ikkita funktsiya yig'indisi uchun yuqorida qilganimiz kabi shunga o'xshash isbotni qilishimiz mumkin. Biroq, biz buni ma'lum miqdorda isbotlaganimiz va buni isbotlaganimizdan ham foydalanishimiz mumkin. 1.

\[\begin{align*}\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - g\left( x \right)} \right] & = \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) + \left( { - 1} \right)g\left( x \right)} \right]\\ & = \mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} \left( { - 1} \right)g\left( x \right)\hspace{0.25in}{\mbox{by first part of }}{\bf{2}}{\mbox{.}}\\ & = \mathop {\lim }\limits_{x \to a} f\left( x \right) + \left( { - 1} \right)\mathop {\lim }\limits_{x \to a} g\left( x \right)\hspace{0.25in}{\mbox{by }}{\bf{1}}{\mbox{.}}\\ & = K + \left( { - 1} \right)L\\ & = K - L\end{align*}\]

Bu biroz qiyin. Birinchidan, shuni ta'kidlaymizki, \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = K\) va \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = L\) biz foydalanishingiz mumkin 2 va 7 quyidagi ikkita chegarani isbotlash uchun.

\[\begin{align*}\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - K} \right] & = \mathop {\lim }\limits_{x \to a} f\left( x \right) - \mathop {\lim }\limits_{x \to a} K = K - K = 0\\ \mathop {\lim }\limits_{x \to a} \left[ {g\left( x \right) - L} \right] & = \mathop {\lim }\limits_{x \to a} g\left( x \right) - \mathop {\lim }\limits_{x \to a} L = L - L = 0\end{align*}\]

Keling, \(\varepsilon > 0\). Keyin \({\delta _{\,1}} > 0\) va \({\delta _2} > 0\) borki,

\[\begin{align*}\left| {\left( {f\left( x \right) - K} \right) - 0} \right| & < \sqrt \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,1}}\\ \left| {\left( {g\left( x \right) - L} \right) - 0} \right| & < \sqrt \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,2}}\end{align*}\]

\(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) ni tanlang. Agar \(0 < \left| {x - a} \right| < \delta \) biz olamiz,

\[\begin{align*}\left| {\left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] - 0} \right| & = \left| {f\left( x \right) - K} \right|\left| {g\left( x \right) - L} \right|\\ & < \sqrt \varepsilon \sqrt \varepsilon \\ & = \varepsilon \end{align*}\]

Shunday qilib, biz buni isbotlashga muvaffaq bo'ldik,

\[\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] = 0\]

Ko'rinishidan, bu biz isbotlamoqchi bo'lgan narsaga hech qanday aloqasi yo'q, lekin birozdan keyin ko'rib turganingizdek, bu kerak.

Haqiqiy dalilga kirishdan oldin 3 keling, bir oz algebra qilaylik. Birinchidan, quyidagi mahsulotni kengaytiring.

\[\left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] = f\left( x \right)g\left( x \right) - Lf\left( x \right) - Kg\left( x \right) + KL\]

Buni qayta tartibga solish ikki funktsiyaning mahsulotini yozishning quyidagi usulini beradi.

\[f\left( x \right)g\left( x \right) = \left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] + Lf\left( x \right) + Kg\left( x \right) - KL\]

Shu bilan biz endi isbotlashni davom ettirishimiz mumkin 3.

\[\begin{align*}\mathop {\lim }\limits_{x \to a} f\left( x \right)g\left( x \right) & = \mathop {\lim }\limits_{x \to a} \left[ {\left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] + Lf\left( x \right) + Kg\left( x \right) - KL} \right]\\ & = \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] + \mathop {\lim }\limits_{x \to a} Lf\left( x \right) + \mathop {\lim }\limits_{x \to a} Kg\left( x \right) - \mathop {\lim }\limits_{x \to a} KL\\ & = 0 + \mathop {\lim }\limits_{x \to a} Lf\left( x \right) + \mathop {\lim }\limits_{x \to a} Kg\left( x \right) - \mathop {\lim }\limits_{x \to a} KL\hspace{0.5in}\hspace{0.5in}\\ & = LK + KL - KL\\ & = LK\end{align*}\]

Haqiqatan ham, boshlashdan oldin bajarishingiz kerak bo'lgan barcha qadamlarni ko'rganingizdan so'ng, juda oddiy dalil. Ikkinchi bosqichda mulkdan bir nechta foydalanish amalga oshirildi 2. Uchinchi bosqichda biz dastlab isbotlagan chegaradan foydalandik. To'rtinchi bosqichda biz xususiyatlardan foydalandik 1 va 7. Nihoyat, biz biroz soddalashtirdik.

Bu ham biroz qiyin. Birinchidan, biz isbotlashdan boshlaymiz,

\[\mathop {\lim }\limits_{x \to a} \frac{1}{{g\left( x \right)}} = \frac{1}{L}\]

\(\varepsilon > 0\) ga ruxsat bering. Bu bizga darhol kerak emas, lekin bu dalillar har doim ushbu bayonot bilan boshlanadi. Endi, chunki \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = L\) \({\delta _{\,1}} > 0\) mavjud, shundayki,

\[\left| {g\left( x \right) - L} \right| < \frac{{\left| L \right|}}{2}\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,1}}\]

Endi, bizda \(0 < \left| {x - a} \right| < {\delta _{\,1}}\) bor deb faraz qilsak,

\[\begin{align*}\left| L \right| & = \left| {L - g\left( x \right) + g\left( x \right)} \right| & \hspace{0.5in} & {\mbox{just adding zero to }}L\\ & \le \left| {L - g\left( x \right)} \right| + \left| {g\left( x \right)} \right| & \hspace{0.5in} & {\mbox{using the triangle inequality}}\\ & = \left| {g\left( x \right) - L} \right| + \left| {g\left( x \right)} \right| & \hspace{0.5in} & \left| {L - g\left( x \right)} \right| = \left| {g\left( x \right) - L} \right|\\ & < \frac{{\left| L \right|}}{2} + \left| {g\left( x \right)} \right| & \hspace{0.5in} & {\mbox{assuming that }}0 < \left| {x - a} \right| < {\delta _1}\end{align*}\]

Buni qayta tartibga solish,

\[\left| L \right| < \frac{{\left| L \right|}}{2} + \left| {g\left( x \right)} \right|\hspace{0.25in} \Rightarrow \hspace{0.25in}\frac{{\left| L \right|}}{2} < \left| {g\left( x \right)} \right|\hspace{0.5in} \Rightarrow \hspace{0.25in}\frac{1}{{\left| {g\left( x \right)} \right|}} < \frac{2}{{\left| L \right|}}\]

Endi \({\delta _2} > 0\) ham borki,

\[\left| {g\left( x \right) - L} \right| < \frac{{{{\left| L \right|}^2}}}{2}\varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,2}}\]

\(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) ni tanlang. Agar bizda \(0 < \left| {x - a} \right| < \delta \) bo'lsa,

\[\begin{align*}\left| {\frac{1}{{g\left( x \right)}} - \frac{1}{L}} \right| & = \left| {\frac{{L - g\left( x \right)}}{{Lg\left( x \right)}}} \right| & \hspace{0.5in} & {\mbox{common denominators}}\\ & = \frac{1}{{\left| {Lg\left( x \right)} \right|}}\left| {L - g\left( x \right)} \right| & \hspace{0.25in} & {\mbox{doing a little rewriting}}\\ & = \frac{1}{{\left| L \right|}}\frac{1}{{\left| {g\left( x \right)} \right|}}\left| {g\left( x \right) - L} \right| & \hspace{0.25in} & {\mbox{doing a little more rewriting}}\\ & < \frac{1}{{\left| L \right|}}\frac{2}{{\left| L \right|}}\left| {g\left( x \right) - L} \right| & \hspace{0.25in} & {\mbox{assuming that }}0 < \left| {x - a} \right| < \delta \le {\delta _1}\\ & < \frac{2}{{{{\left| L \right|}^2}}}\frac{{{{\left| L \right|}^2}}}{2}\varepsilon & \hspace{0.5in} & {\mbox{assuming that }}0 < \left| {x - a} \right| < \delta \le {\delta _2}\\ & = \varepsilon \end{align*}\]

Endi biz \(\mathop {\lim }\limits_{x \to a} \frac{1}{{g\left( x \right)}} = \frac{1}{L}\) ni isbotlaganimizdan so'ng, umumiy haqiqat oson.

\[\begin{align*}\mathop {\lim }\limits_{x \to a} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] & = \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right)\frac{1}{{g\left( x \right)}}} \right]\\ & = \mathop {\lim }\limits_{x \to a} f\left( x \right)\mathop {\lim }\limits_{x \to a} \frac{1}{{g\left( x \right)}}\hspace{0.5in}{\mbox{using property }}{\bf{3}}{\mbox{.}}\\ & = K\frac{1}{L} = \frac{K}{L}\end{align*}\]

Ta'kidlanganidek, biz faqat isbotlaymiz 5 butun sonli darajalar uchun. Bu shuningdek, induksiya orqali isbotlashni o'z ichiga oladi, shuning uchun agar siz induksiya dalillari bilan tanish bo'lmasangiz, bu dalilni o'tkazib yuborishingiz mumkin.

Shunday qilib, biz isbotlaymiz,

\[\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^n} = {\left[ {\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right]^n} = {K^n},\hspace{0.5in}n \ge 2,\,\,n{\mbox{ is an integer}}{\mbox{.}}\]

\(n = 2\) uchun bizda maxsus mulk holatidan boshqa narsa yo'q 3.

\[\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^2} = \mathop {\lim }\limits_{x \to a} f\left( x \right)f\left( x \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right)\mathop {\lim }\limits_{x \to a} f\left( x \right) = KK = {K^2}\]

So, 5 \(n = 2\) uchun tasdiqlangan. Endi taxmin qiling 5 \(n - 1\) yoki \(\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^{n - 1}} = {K^{n - 1}}\) uchun to'g'ri. Keyin yana mulkdan foydalaning 3 bizda bor,

\[\begin{align*}\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^n} & = \mathop {\lim }\limits_{x \to a} \left( {{{\left[ {f\left( x \right)} \right]}^{n - 1}}f\left( x \right)} \right)\\ & = \mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^{n - 1}}\mathop {\lim }\limits_{x \to a} f\left( x \right)\\ & = {K^{n - 1}}K\\ & = {K^n}\end{align*}\]

da ta'kidlanganidek Cheklash xususiyatlari bo'lim, bu to'liq versiyasining maxsus holatidan boshqa narsa emas 5 va dalil u erda berilgan va bu erda dalil berilmaydi.

Bu oddiy dalil. Belgilashni biroz osonlashtirish uchun \(f\left( x \right) = x\) ni aniqlasak, bizdan \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = a\) ekanligini isbotlash talab qilinadi.

\(\varepsilon > 0\) va \(\delta = \varepsilon \) bo'lsin. Keyin, agar bizda \(0 < \left| {x - a} \right| < \delta = \varepsilon \) bo'lsa,

\[\left| {f\left( x \right) - a} \right| = \left| {x - a} \right| < \delta = \varepsilon \]

Shunday qilib, biz \(\mathop {\lim }\limits_{x \to a} x = a\) ekanligini isbotladik.

Bu faqat mulkning alohida holati 5 \(f\left( x \right) = x\) bilan va shuning uchun biz buni bu erda isbotlamaymiz.

Biz bu erda \(\mathop {\lim }\limits_{x \to c} \left[ {f\left( x \right) + g\left( x \right)} \right] = \infty \) ni isbotlaymiz. \(\mathop {\lim }\limits_{x \to c} \left[ {f\left( x \right) - g\left( x \right)} \right] = \infty \) isboti deyarli bir xil va sizga qoldiriladi.

\(M > 0\) bo'lsin, chunki biz \(\mathop {\lim }\limits_{x \to c} f\left( x \right) = \infty \) ni bilamiz, chunki \({\delta _{\,1}} > 0\) bor, agar \(0 < \left| {x - c} \right| < {\delta _{\,1}}\) bizda bo'lsa,

\[f\left( x \right) > M - L + 1\]

Bundan tashqari, biz \(\mathop {\lim }\limits_{x \to c} g\left( x \right) = L\) ni bilganimiz uchun \({\delta _{\,2}} > 0\) mavjud, agar \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) bizda mavjud bo'lsa,

\[0 < \left| {g\left( x \right) - L} \right| < 1\hspace{0.25in}\,\, \to \hspace{0.25in} - 1 < g\left( x \right) - L < 1\hspace{0.25in} \to \,\,\,\,\,\,\,L - 1 < g\left( x \right) < L + 1\]

Endi, \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) ga ruxsat bering va agar \(0 < \left| {x - c} \right| < \delta \) yuqoridagi bayonotlardan biz ikkalasiga ham ega bo'lishini bilsak,

\[f\left( x \right) > M - L + 1\hspace{0.5in}\hspace{0.25in}g\left( x \right) > L - 1\]

Bu bizga beradi,

\[\begin{align*}f\left( x \right) + g\left( x \right) & > M - L + 1 + L - 1\\ & = M\hspace{0.5in}\hspace{0.25in} \Rightarrow \hspace{0.5in}f\left( x \right) + g\left( x \right) > M\end{align*}\]

Shunday qilib, biz \(\mathop {\lim }\limits_{x \to c} \left[ {f\left( x \right) + g\left( x \right)} \right] = \infty \) ekanligini isbotladik.

\(M > 0\) bo'lsin, chunki biz \(\mathop {\lim }\limits_{x \to c} f\left( x \right) = \infty \) ni bilamiz, chunki \({\delta _{\,1}} > 0\) bor, agar \(0 < \left| {x - c} \right| < {\delta _{\,1}}\) bizda bo'lsa,

\[f\left( x \right) > \frac{{2M}}{L}\]

Bundan tashqari, biz \(\mathop {\lim }\limits_{x \to c} g\left( x \right) = L\) ni bilganimiz uchun \({\delta _{\,2}} > 0\) mavjud, agar \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) bizda mavjud bo'lsa,

\[0 < \left| {g\left( x \right) - L} \right| < \frac{L}{2}\hspace{0.25in} \to \hspace{0.25in} - \frac{L}{2} < g\left( x \right) - L < \frac{L}{2}\hspace{0.25in} \to \hspace{0.25in} \frac{L}{2} < g\left( x \right) < \frac{{3L}}{2}\]

E'tibor bering, chunki yuqoridagi birinchi tengsizlikda \(L > 0\) ni tanlagan \(\frac{L}{2}\) to'g'ri tanlovdir, chunki u chegara ta'rifi talab qilganidek ijobiy bo'ladi.

Endi, \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) bo'lsin va agar \(0 < \left| {x - c} \right| < \delta \) yuqoridagi bayonotlardan biz ikkalasiga ham ega bo'lishini bilsak,

\[f\left( x \right) > \frac{{2M}}{L}\hspace{0.5in}\hspace{0.25in}g\left( x \right) > \frac{L}{2}\]

Bu bizga beradi,

\[\begin{align*}f\left( x \right)g\left( x \right) & > \left( {\frac{{2M}}{L}} \right)\left( {\frac{L}{2}} \right)\\ & = M\hspace{0.5in}\hspace{0.25in} \Rightarrow \hspace{0.5in}f\left( x \right)g\left( x \right) > M\end{align*}\]

Shunday qilib, biz \(\mathop {\lim }\limits_{x \to c} f\left( x \right)g\left( x \right) = \infty \) ekanligini isbotladik.

\(M > 0\) bo'lsin, chunki biz \(\mathop {\lim }\limits_{x \to c} f\left( x \right) = \infty \) ni bilamiz, chunki \({\delta _{\,1}} > 0\) bor, agar \(0 < \left| {x - c} \right| < {\delta _{\,1}}\) bizda bo'lsa,

\[f\left( x \right) > \frac{{ - 2M}}{L}\]

E'tibor bering, chunki \(L < 0\) bu erda bizda \(\frac{{ - 2M}}{L} > 0\) bo'ladi.

Keyinchalik, biz \(\mathop {\lim }\limits_{x \to c} g\left( x \right) = L\) ni bilganimiz uchun \({\delta _{\,2}} > 0\) mavjud, agar \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) bizda mavjud bo'lsa,

\[0 < \left| {g\left( x \right) - L} \right| < - \frac{L}{2}\hspace{0.25in}\,\, \to \hspace{0.25in}\frac{L}{2} < g\left( x \right) - L < - \frac{L}{2}\hspace{0.25in} \to \hspace{0.25in} \frac{{3L}}{2} < g\left( x \right) < \frac{L}{2}\]

Shunga qaramay, biz \(L < 0\) ekanligini bilganimiz uchun bizda \( - \frac{L}{2} > 0\) bo'ladi. Bundan tashqari, tez orada aniq bo'ladigan sabablarga ko'ra, yakuniy tengsizlikni minus belgisiga ko'paytiring,

\[\, - \frac{L}{2} < - g\left( x \right) < - \frac{{3L}}{2}\]

Endi \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) ga ruxsat bering va agar \(0 < \left| {x - c} \right| < \delta \) yuqoridagi bayonotlardan biz ikkalasiga ham ega bo'lishini bilsak,

\[f\left( x \right) > \frac{{ - 2M}}{L}\hspace{0.5in}\hspace{0.25in} - g\left( x \right) > - \frac{L}{2}\]

Bu bizga beradi,

\[\begin{align*} - f\left( x \right)g\left( x \right) & = f\left( x \right)\left[ { - g\left( x \right)} \right]\\ & > \left( { - \frac{{2M}}{L}} \right)\left( { - \frac{L}{2}} \right)\\ & = M\hspace{0.5in}\hspace{0.25in} \Rightarrow \hspace{0.5in} - f\left( x \right)g\left( x \right) > M\end{align*}\]

Bu bizga kerak bo'lgan narsa emasdek tuyulishi mumkin, ammo buni minus belgisiga ko'paytirsak,

\[f\left( x \right)g\left( x \right) < - M\]

va biz dastlab \(M > 0\) ni tanlaganimiz uchun endi \(\mathop {\lim }\limits_{x \to c} f\left( x \right)g\left( x \right) = - \infty \) ekanligini isbotladik.

Buni uchta holatda qilishimiz kerak. Eng oson holatdan boshlaylik.

1-holat: \(L = 0\)

\(\varepsilon > 0\) bo'lsin, chunki biz \(\mathop {\lim }\limits_{x \to c} f\left( x \right) = \infty \) borligini bilamiz, chunki \({\delta _{\,1}} > 0\) bor, agar \(0 < \left| {x - c} \right| < {\delta _{\,1}}\) bizda bo'lsa,

\[f\left( x \right) > \frac{1}{{\sqrt \varepsilon }} > 0\]

Keyinchalik, biz \(\mathop {\lim }\limits_{x \to c} g\left( x \right) = 0\) ni bilganimiz uchun \({\delta _{\,2}} > 0\) mavjud, agar \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) bizda mavjud bo'lsa,

\[0 < \left| {g\left( x \right)} \right| < \sqrt \varepsilon \]

Endi, \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) bo'lsin va agar \(0 < \left| {x - c} \right| < \delta \) yuqoridagi bayonotlardan biz ikkalasiga ham ega bo'lishini bilsak,

\[f\left( x \right) > \frac{1}{{\sqrt \varepsilon }}\hspace{0.5in}\hspace{0.25in}\left| {g\left( x \right)} \right| < \sqrt \varepsilon \]

Bu bizga beradi,

\[\left| {\frac{{g\left( x \right)}}{{f\left( x \right)}}} \right| = \frac{{\left| {g\left( x \right)} \right|}}{{f\left( x \right)}} < \frac{{\sqrt \varepsilon }}{{f\left( x \right)}} < \frac{{\sqrt \varepsilon }}{{{}^{1}/{}_{{\sqrt \varepsilon }}}} = \varepsilon \]

Ikkinchi bosqichda biz \(f\left( x \right)\) dan mutlaq qiymat satrlarini olib tashlashimiz mumkin, chunki biz bu ijobiy ekanligini bilamiz.

Shunday qilib, biz \(L = 0\) bo'lsa, \(\mathop {\lim }\limits_{x \to c} \frac{{g\left( x \right)}}{{f\left( x \right)}} = 0\) ekanligini isbotladik.

2-holat: \(L > 0\)

\(\varepsilon > 0\) bo'lsin, chunki biz \(\mathop {\lim }\limits_{x \to c} f\left( x \right) = \infty \) ni bilamiz, chunki \({\delta _{\,1}} > 0\) bor, agar \(0 < \left| {x - c} \right| < {\delta _{\,1}}\) bizda bo'lsa,

\[f\left( x \right) > \frac{{3L}}{{2\varepsilon }} > 0\]

Keyinchalik, biz \(\mathop {\lim }\limits_{x \to c} g\left( x \right) = L\) ni bilganimiz uchun \({\delta _{\,2}} > 0\) mavjud, agar \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) bizda mavjud bo'lsa,

\[0 < \left| {g\left( x \right) - L} \right| < \frac{L}{2}\hspace{0.25in}\, \to \,\,\,\,\,\,\, - \frac{L}{2} < g\left( x \right) - L < \frac{L}{2}\hspace{0.25in}\, \to \hspace{0.25in}\,\frac{L}{2} < g\left( x \right) < \frac{{3L}}{2}\]

Bundan tashqari, biz \(L > 0\) deb taxmin qilganimiz sababli, \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) uchun bizda \(g\left( x \right) > 0\) bor deb taxmin qilish xavfsiz.

Endi, \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) ga ruxsat bering va agar \(0 < \left| {x - c} \right| < \delta \) yuqoridagi bayonotlardan biz ikkalasiga ham ega bo'lishini bilsak,

\[f\left( x \right) > \frac{{3L}}{{2\varepsilon }}\hspace{0.5in}\hspace{0.25in}g\left( x \right) < \frac{{3L}}{2}\]

Bu bizga beradi,

\[\left| {\frac{{g\left( x \right)}}{{f\left( x \right)}}} \right| = \frac{{g\left( x \right)}}{{f\left( x \right)}} < \frac{{{}^{{3L}}/{}_{2}}}{{\left| {f\left( x \right)} \right|}} < \frac{{{}^{{3L}}/{}_{2}}}{{{}^{{3L}}/{}_{{2\varepsilon }}}} = \varepsilon \]

Ikkinchi bosqichda biz mutlaq qiymat satrlarini olib tashlashimiz mumkin, chunki biz ikkala funktsiyaning ijobiy ekanligini bilamiz yoki ishonch bilan taxmin qilishimiz mumkin (yuqorida aytib o'tilganidek).

Shunday qilib, biz \(\mathop {\lim }\limits_{x \to c} \frac{{g\left( x \right)}}{{f\left( x \right)}} = 0\) ekanligini isbotladik if \(L > 0\).

3-holat: \(L < 0\).

\(\varepsilon > 0\) bo'lsin, chunki biz \(\mathop {\lim }\limits_{x \to c} f\left( x \right) = \infty \) ni bilamiz, chunki \({\delta _{\,1}} > 0\) bor, agar \(0 < \left| {x - c} \right| < {\delta _{\,1}}\) bizda bo'lsa,

\[f\left( x \right) > \frac{{ - 3L}}{{2\varepsilon }} > 0\]

Keyinchalik, biz \(\mathop {\lim }\limits_{x \to c} g\left( x \right) = L\) ni bilganimiz uchun \({\delta _{\,2}} > 0\) mavjud, agar \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) bizda mavjud bo'lsa,

\[0 < \left| {g\left( x \right) - L} \right| < - \frac{L}{2}\hspace{0.25in}\, \to \,\,\,\,\,\,\,\frac{L}{2} < g\left( x \right) - L < - \frac{L}{2}\hspace{0.25in}\, \to \hspace{0.25in}\,\frac{{3L}}{2} < g\left( x \right) < \frac{L}{2}\]

Keyin, buni olish uchun salbiy belgi bo'lishini ko'paytiring,

\[ - \frac{L}{2} < - g\left( x \right) < - \frac{{3L}}{2}\]

Bundan tashqari, biz \(L < 0\) deb taxmin qilganimiz sababli, \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) uchun bizda \(g\left( x \right) < 0\) bor deb taxmin qilish xavfsiz.

Endi, \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) bo'lsin va agar \(0 < \left| {x - c} \right| < \delta \) yuqoridagi bayonotlardan biz ikkalasiga ham ega bo'lishini bilsak,

\[f\left( x \right) > \frac{{ - 3L}}{{2\varepsilon }}\hspace{0.5in}\hspace{0.25in} - g\left( x \right) < \frac{{ - 3L}}{2}\]

Bu bizga beradi,

\[\left| {\frac{{g\left( x \right)}}{{f\left( x \right)}}} \right| = \frac{{ - g\left( x \right)}}{{f\left( x \right)}} < \frac{{{}^{{ - 3L}}/{}_{2}}}{{\left| {f\left( x \right)} \right|}} < \frac{{{}^{{ - 3L}}/{}_{2}}}{{{}^{{ - 3L}}/{}_{{2\varepsilon }}}} = \varepsilon \]

Ikkinchi bosqichda biz mutlaq qiymat satrlarini salbiyga qo'shish orqali olib tashlashimiz mumkin, chunki biz \(f\left( x \right) > 0\) ekanligini bilamiz va \(g\left( x \right) < 0\) ni ishonch bilan qabul qilishimiz mumkin (yuqorida aytib o'tilganidek).

Shunday qilib, biz \(\mathop {\lim }\limits_{x \to c} \frac{{g\left( x \right)}}{{f\left( x \right)}} = 0\) ekanligini isbotladik if \(L < 0\).

Bu aslida juda oddiy dalil, lekin biz uchta alohida ishni bajarishimiz kerak.

1-holat: \(c > 0\) deb faraz qiling. Keyin \(\varepsilon > 0\) va aniqlang

\[M = \sqrt[r]{{\frac{c}{\varepsilon }}}\]

E'tibor bering, \(c\) va \(\varepsilon \) ikkalasi ham ijobiy bo'lgani uchun biz bu ildiz mavjudligini bilamiz. Endi bizda bor deb taxmin qiling

\[x > M = \sqrt[r]{{\frac{c}{\varepsilon }}}\]

Ushbu taxminni hisobga olsak,

\[\begin{align*}x > \sqrt[r]{{\frac{c}{\varepsilon }}} & & \\ {x^r} & > \frac{c}{\varepsilon }& \hspace{0.5in} & {\mbox{get rid of the root}}\\ \frac{c}{{{x^r}}} & < \varepsilon & \hspace{0.5in} & {\mbox{rearrange things a little}}\\ \left| {\frac{c}{{{x^r}}} - 0} \right| & < \varepsilon & \hspace{0.5in} & {\mbox{everything is positive so we can add absolute value bars}}\end{align*}\]

Shunday qilib, \(c > 0\) bilan biz \(\mathop {\lim }\limits_{x \to \infty } \frac{c}{{{x^r}}} = 0\) ekanligini isbotladik.

2-holat: \(c = 0\) deb faraz qiling. Bu erda biz qilishimiz kerak bo'lgan yagona narsa:

\[\mathop {\lim }\limits_{x \to \infty } \frac{c}{{{x^r}}} = \mathop {\lim }\limits_{x \to \infty } \frac{0}{{{x^r}}} = \mathop {\lim }\limits_{x \to \infty } 0 = 0\]

3-holat: Nihoyat, \(c < 0\) deb faraz qiling. Bunday holda, biz \(c = - k\) deb yozishimiz mumkin, bu erda \(k > 0\). Keyin 1-holatidan foydalanib va biz doimiylarni chegaradan chiqarib tashlashimiz mumkin,

\[\mathop {\lim }\limits_{x \to \infty } \frac{c}{{{x^r}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{ - k}}{{{x^r}}} = - \mathop {\lim }\limits_{x \to \infty } \frac{k}{{{x^r}}} = - 0 = 0\]

Bu dalilga juda o'xshaydi 1 shuning uchun biz faqat birinchi ishni qilamiz (eng qiyin bo'lgani uchun) va qolgan ikkita ishni isbotlash uchun sizga qoldiramiz.

1-holat: \(c > 0\) deb faraz qiling. Keyin \(\varepsilon > 0\) va aniqlang

\[N = - \sqrt[r]{{\frac{c}{\varepsilon }}}\]

E'tibor bering, \(c\) va \(\varepsilon \) ikkalasi ham ijobiy bo'lgani uchun biz bu ildiz mavjudligini bilamiz. Endi bizda bor deb taxmin qiling

\[x < N = - \sqrt[r]{{\frac{c}{\varepsilon }}}\]

E'tibor bering, bu taxmin \(x\) salbiy bo'lishini ham aytadi. Bizda bu taxminni keltiring,

\[\begin{align*}x & < - \sqrt[r]{{\frac{c}{\varepsilon }}} & & \\ \left| x \right| & > \left| {\sqrt[r]{{\frac{c}{\varepsilon }}}} \right| & \hspace{0.25in} & {\mbox{take absolute value of both sides}}\\ \left| {{x^r}} \right| & > \left| {\frac{c}{\varepsilon }} \right| & \hspace{0.25in} & {\mbox{get rid of the root}}\\ \left| {\frac{c}{{{x^r}}}} \right| & < \left| \varepsilon \right| = \varepsilon & \hspace{0.25in} & {\mbox{rearrange things a little and use the fact that }}\varepsilon > 0\\ \left| {\frac{c}{{{x^r}}} - 0} \right| & < \varepsilon & \hspace{0.25in} & {\mbox{rewrite things a little}}\end{align*}\]

Shunday qilib, \(c > 0\) bilan biz \(\mathop {\lim }\limits_{x \to - \infty } \frac{c}{{{x^r}}} = 0\) ekanligini isbotladik. E'tibor bering, bu erda asosiy farq shundaki, biz minus belgisi bilan shug'ullanish uchun birinchi navbatda mutlaq qiymatni olishimiz kerak. Ikkala tomon ham manfiy bo'lgani uchun biz bilamizki, ikkala tomonning mutlaq qiymatini olganda, tengsizlikning yo'nalishi ham o'zgarishi kerak.

2-holat, 3-holat: Yuqorida ta'kidlanganidek, ular birinchi dalildagi tegishli holatlarning isboti bilan bir xil va shuning uchun bu erda olib tashlangan.

Biz buni polinomlar bilan bog'liq bo'lgan ushbu bo'limda ishlagan masalalarga bir xil tarzda isbotlaymiz. Biz birinchi navbatda polinomdan \({a_n}{x^n}\) ni ajratamiz va keyin katta hajmdan foydalanamiz. 1-fakt (buni biz yuqorida isbotladik) va chegaralarning asosiy xossalari.

\[\begin{align*}\mathop {\lim }\limits_{x \to \infty } p\left( x \right) & = \mathop {\lim }\limits_{x \to \infty } \left( {{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + \cdots + {a_1}x + {a_0}} \right)\\ & = \mathop {\lim }\limits_{x \to \infty } \left[ {{a_n}{x^n}\left( {1 + \frac{{{a_{n - 1}}}}{{{a_n}x}} + \cdots + \frac{{{a_1}}}{{{a_n}{x^{n - 1}}}} + \frac{{{a_0}}}{{{a_n}{x^n}}}} \right)} \right]\end{align*}\]

Endi, aniqki, ikkinchi hadning chegarasi bitta va birinchi hadning chegarasi \({a_n}\) belgisiga qarab \(\infty \) yoki \( - \infty \) bo'ladi. Shuning uchun tomonidan Faktlar Cheksiz limitlar bo'limidan biz butun ko'phadning chegarasi birinchi hadning chegarasi bilan bir xil bo'lishini ko'rishimiz mumkin yoki,

\[\mathop {\lim }\limits_{x \to \infty } p\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \left( {{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + \cdots + {a_1}x + {a_0}} \right) = \mathop {\lim }\limits_{x \to \infty } {a_n}{x^n}\]

Bu qismning isboti birinchi qismning isboti bilan tom ma'noda bir xil, bundan mustasno, barcha \(\infty \) lar \( - \,\infty \) ga o'zgartirilgan va shuning uchun bu erda olib tashlangan.

\(\varepsilon > 0\) bo'lsin, keyin biz \(\delta > 0\) mavjudligini ko'rsatishimiz kerak, shunday qilib,

\[\left| {f\left( {g\left( x \right)} \right) - f\left( b \right)} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta \]

Keling, \(f\left( x \right)\) \(x = b\) da uzluksiz ekanligidan boshlaylik. Eslatib o'tamiz, bu \(\mathop {\lim }\limits_{x \to b} f\left( x \right) = f\left( b \right)\) va shuning uchun \({\delta _{\,1}} > 0\) bo'lishi kerak degan ma'noni anglatadi,

\[\left| {f\left( x \right) - f\left( b \right)} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - b} \right| < {\delta _{\,1}}\]

Keling, \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = b\) ni eslaylik. Bu \(\delta > 0\) bo'lishi kerakligini anglatadi, shuning uchun,

\[\left| {g\left( x \right) - b} \right| < {\delta _{\,1}}\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta \]

Ammo bularning barchasi biz tugatganimizni anglatadi.

Keling, xulosa qilaylik. Avval \(0 < \left| {x - a} \right| < \delta \) deb faraz qiling. Bu bizga shuni aytadiki,

\[\left| {g\left( x \right) - b} \right| < {\delta _{\,1}}\]

Ammo, agar \(0 < \left| {x - b} \right| < {\delta _{\,1}}\) bo'lsa, bizda \(\left| {f\left( x \right) - f\left( b \right)} \right| < \varepsilon \) bo'lishi kerakligini ham bilamiz. Bu bizga shuni anglatadiki, agar raqam \(b\) dan \({\delta _{\,1}}\) masofasida bo'lsa, biz bu raqamni \(f\left( x \right)\) ga ulashimiz mumkin va biz \(f\left( b \right)\) \(\varepsilon \) masofasida bo'lamiz.

Shunday qilib, \[\left| {g\left( x \right) - b} \right| < {\delta _{\,1}}\] bizga \(g\left( x \right)\) \(b\) ning \({\delta _{\,1}}\) masofasida ekanligini aytadi va shuning uchun uni \(f\left( x \right)\) ga ulab olsak, biz quyidagilarni olamiz:

\[\left| {f\left( {g\left( x \right)} \right) - f\left( b \right)} \right| < \varepsilon \]

va bu biz ko'rsatmoqchi bo'lgan narsadir.