2.10-bo'lim : Limitning ta'rifi
2. Quyidagi chegarani isbotlash uchun chegara ta’rifidan foydalaning.
\[\mathop {\lim }\limits_{x \to - 1} \left( {x + 7} \right) = 6\]Yechim
Birinchidan, nima ko'rsatishimiz kerakligini yozamiz.
\(\varepsilon > 0\) istalgan raqam bo'lsin. Biz \(\delta > 0\) raqamini topishimiz kerak, shunda
\[\left| {\left( {x + 7} \right) - 6} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - \left( { - 1} \right)} \right| < \delta \]Yoki biroz soddalashtirilgan holda bu shunday bo'ladi:
\[\left| {x + 1} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x + 1} \right| < \delta \]2-qadam
Bu muammo shu nuqtadan boshlab 1-masalaga juda o'xshaydi.
Biz \(\delta \) ni aniqlashimiz kerak, bu bizga 1-qadamdagi bayonotni isbotlash imkonini beradi. Biroq, ikkala tengsizlik ham aynan bir xil mutlaq qiymat bayonotini o'z ichiga oladi, faqat \(\delta = \varepsilon \) ni tanlashimiz kerak.
3-qadam
Shunday qilib, keling, bu ish yoki yo'qligini bilib olaylik.
Avval \(\varepsilon > 0\) istalgan raqam deb faraz qilishdan boshlang va \(\delta = \varepsilon \) ni tanlang. Endi biz taxmin qilishimiz mumkin,
\[0 < \left| {x - \left( { - 1} \right)} \right| < \delta = \varepsilon \hspace{0.5in} \Rightarrow \hspace{0.5in}0 < \left| {x + 1} \right| < \varepsilon \]Bu beradi,
\[\begin{align*}\left| {\left( {x + 7} \right) - 6} \right| & = \left| {x + 1} \right| & & \hspace{0.25in}{\mbox{simplify things up a little}}\\ & < \varepsilon & & \hspace{0.25in}{\mbox{using the information we got by assuming }}\delta = \varepsilon \end{align*}\]Shunday qilib, biz buni ko'rsatdik,
\[\left| {\left( {x + 7} \right) - 6} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - \left( { - 1} \right)} \right| < \varepsilon \]va shuning uchun chegaraning ta'rifi bilan biz buni isbotladik,
\[\mathop {\lim }\limits_{x \to - 1} \left( {x + 7} \right) = 6\]