2.10-bo'lim : Limitning ta'rifi
5. Quyidagi chegarani isbotlash uchun chegara ta’rifidan foydalaning.
\[\mathop {\lim }\limits_{x \to 1} \frac{1}{{{{\left( {x - 1} \right)}^2}}} = \infty \]Yechim
Birinchidan, nima ko'rsatishimiz kerakligini yozamiz.
\(M > 0\) istalgan raqam bo'lsin. Biz \(\delta > 0\) raqamini topishimiz kerak, shunda
\[\frac{1}{{{{\left( {x - 1} \right)}^2}}} > M\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - 1} \right| < \delta \]2-qadam
Keling, yuqoridagi birinchi tengsizlikni biroz qayta yozaylik.
\[\frac{1}{{{{\left( {x - 1} \right)}^2}}} > M\hspace{0.25in}\,\, \to \hspace{0.25in}\,\,\,\,\,{\left( {x - 1} \right)^2} < \frac{1}{M}\hspace{0.25in}\,\,\,\,\,\,\, \to \hspace{0.25in}\,\,\,\,\left| {x - 1} \right| < \frac{1}{{\sqrt M }}\]Bundan biz \(\delta = \frac{1}{{\sqrt M }}\) ni tanlashimiz mumkin.
3-qadam
Shunday qilib, keling, bu ish yoki yo'qligini bilib olaylik.
Biz \(M > 0\) istalgan raqam deb faraz qilishdan boshlaymiz va \(\delta = \frac{1}{{\sqrt M }}\) ni tanlaymiz. Endi biz taxmin qilishimiz mumkin,
\[0 < \left| {x - 1} \right| < \delta = \frac{1}{{\sqrt M }}\hspace{0.5in} \Rightarrow \hspace{0.5in}0 < \left| {x - 1} \right| < \frac{1}{{\sqrt M }}\]Shunday qilib, agar biz ikkinchi tengsizlikdan boshlasak,
\[\begin{align*}\left| {x - 1} \right| & < \frac{1}{{\sqrt M }} & & \\ & {\left| {x - 1} \right|^2} < \frac{1}{M} & & \hspace{0.25in}{\mbox{squaring both sides}}\\ & {\left( {x - 1} \right)^2} < \frac{1}{M} & & \hspace{0.25in}\,\,{\mbox{because }}{\left| {x - 1} \right|^2} = {\left( {x - 1} \right)^2}\\ & \frac{1}{{{{\left( {x - 1} \right)}^2}}} > M & & \hspace{0.25in}{\mbox{ rewriting things a little bit}}\end{align*}\]Shunday qilib, biz buni ko'rsatdik,
\[\frac{1}{{{{\left( {x - 1} \right)}^2}}} > M\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - 1} \right| < \frac{1}{{\sqrt M }}\]va shuning uchun chegaraning ta'rifi bilan biz buni isbotladik,
\[\mathop {\lim }\limits_{x \to 1} \frac{1}{{{{\left( {x - 1} \right)}^2}}} = \infty \]