2.10-bo'lim : Limitning ta'rifi
7. Quyidagi chegarani isbotlash uchun chegara ta'rifidan foydalaning.
\[\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}} = 0\]Yechim
Birinchidan, nima ko'rsatishimiz kerakligini yozamiz.
\(\varepsilon > 0\) istalgan raqam bo'lsin. Biz \(M > 0\) raqamini topishimiz kerak, shunda
\[\left| {\frac{1}{{{x^2}}} - 0} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}x > M\]Yoki biroz soddalashtirilgan holda bu shunday bo'ladi:
\[\left| {\frac{1}{{{x^2}}}} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}x > M\]2-qadam
Chapdagi tengsizlikdan boshlaylik va unga biroz qayta yozamiz.
\[\left| {\frac{1}{{{x^2}}}} \right| < \varepsilon \hspace{0.25in} \to \hspace{0.25in}\frac{1}{{{{\left| x \right|}^2}}} < \varepsilon \hspace{0.25in}\, \to \hspace{0.25in}{\left| x \right|^2} > \frac{1}{\varepsilon }\hspace{0.25in} \to \hspace{0.25in}\,\left| x \right| > \frac{1}{{\sqrt \varepsilon }}\]Bundan biz \(M = \frac{1}{{\sqrt \varepsilon }}\) ni tanlashimiz mumkin
3-qadam
Shunday qilib, keling, bu ish yoki yo'qligini bilib olaylik.
Avval \(\varepsilon > 0\) istalgan raqam deb faraz qilishdan boshlang va \(M = \frac{1}{{\sqrt \varepsilon }}\) ni tanlang. Endi biz taxmin qilishimiz mumkin,
\[x > \frac{1}{{\sqrt \varepsilon }}\]Ushbu tengsizlikdan boshlab biz olamiz,
\[\begin{align*}x & > \frac{1}{{\sqrt \varepsilon }} & & \\ & \frac{1}{x} < \sqrt \varepsilon & & \hspace{0.25in}{\mbox{do a little rewrite}}\\ & \frac{1}{{{x^2}}} < \varepsilon & & \hspace{0.25in}{\mbox{square both sides}}\\ & \left| {\frac{1}{{{x^2}}}} \right| < \varepsilon & & \hspace{0.25in}{\mbox{because }}\frac{1}{{{x^2}}} = \left| {\frac{1}{{{x^2}}}} \right|\end{align*}\]Shunday qilib, biz buni ko'rsatdik,
\[\left| {\frac{1}{{{x^2}}} - 0} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}x > M\]va shuning uchun chegaraning ta'rifi bilan biz buni isbotladik,
\[\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}} = 0\]