4.10-bo'lim : L'Hospital qoidalari va noaniq shakllar
10. \(\mathop {\lim }\limits_{y \to {0^ + }} {\left[ {\cos \left( {2y} \right)} \right]^{{}^{1}/{}_{{{y^{\,2}}}}}}\) ni baholash uchun L'Hospital qoidasidan foydalaning.
Yechim
Bu erda e'tibor berish kerak bo'lgan birinchi narsa - bu L'Hospital qoidasiga ruxsat beruvchi shaklda emas. L'Hospital qoidasi faqat ratsional funktsiyalarning ma'lum sinflarida ishlaydi va bu ratsional funktsiya emasligi aniq.
Ammo e'tibor bering, u quyidagi noaniq shaklda,
\[{\mbox{as}}\,\,\,y \to {0^ + }\hspace{0.5in}{\left[ {\cos \left( {2y} \right)} \right]^{{}^{1}/{}_{{{y^{\,2}}}}}} \to {1^\infty }\]va biz ushbu bo'lim uchun eslatmalarda muhokama qilganimizdek, biz buni L'Hospital qoidasi bilan hal qilinishi mumkin bo'lgan muammoga aylantirish uchun ba'zi manipulyatsiyalar qilishimiz mumkin.
2-qadam
Birinchidan, aniqlaymiz,
\[z = {\left[ {\cos \left( {2y} \right)} \right]^{{}^{1}/{}_{{{y^{\,2}}}}}}\]va ikkala tomonning jurnalini oling. Biz ham biroz soddalashtiramiz.
\[\ln z = \ln \left( {{{\left[ {\cos \left( {2y} \right)} \right]}^{{}^{1}/{}_{{{y^{\,2}}}}}}} \right) = \frac{1}{{{y^{\,2}}}}\ln \left[ {\cos \left( {2y} \right)} \right] = \frac{{\ln \left[ {\cos \left( {2y} \right)} \right]}}{{{y^2}}}\]3-qadam
Endi biz chegarani \(y \to {0^ + }\) sifatida qabul qilishimiz mumkin.
\[\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\ln z} \right] = \mathop {\lim }\limits_{y \to {0^ + }} \left[ {\frac{{\ln \left[ {\cos \left( {2y} \right)} \right]}}{{{y^2}}}} \right]\]Davom etishdan oldin, bizda quyidagilar borligini bilib olaylik:
\[{\mbox{as}}\,\,\,y \to {0^ + }\hspace{0.5in}\frac{{\ln \left[ {\cos \left( {2y} \right)} \right]}}{{{y^2}}} \to \frac{{\ln \left( 1 \right)}}{0} = \frac{0}{0}\]va bizda L'Hospital qoidasidan foydalanishimiz mumkin bo'lgan chegaramiz bor.
4-qadam
L'Hospital qoidasini qo'llash,
\[\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\ln z} \right] = \mathop {\lim }\limits_{y \to {0^ + }} \left[ {\frac{{\ln \left[ {\cos \left( {2y} \right)} \right]}}{{{y^2}}}} \right] = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{{}^{{ - 2\sin \left( {2y} \right)}}/{}_{{\cos \left( {2y} \right)}}}}{{2y}} = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{ - \tan \left( {2y} \right)}}{y}\]5-qadam
Endi bizda shunday chegara mavjud:
\[{\mbox{as}}\,\,\,y \to {0^ + }\hspace{0.5in}\frac{{ - \tan \left( {2y} \right)}}{y} \to \frac{0}{0}\]va shuning uchun biz L'Hospital qoidasidan bu borada ham foydalanishimiz mumkin. Shunday qilib,
\[\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\ln z} \right] = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{ - \tan \left( {2y} \right)}}{y} = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{ - 2{{\sec }^2}\left( {2y} \right)}}{1} = - 2\]6-qadam
Endi biz shuni eslashimiz kerak,
\[z = {{\bf{e}}^{\ln z}}\]Bu, o'z navbatida, biz asl chegarani quyidagicha bajarishimiz mumkinligini anglatadi,
\[\mathop {\lim }\limits_{y \to {0^ + }} {\left[ {\cos \left( {2y} \right)} \right]^{{}^{1}/{}_{{{y^{\,2}}}}}} = \mathop {\lim }\limits_{y \to {0^ + }} z = \mathop {\lim }\limits_{y \to {0^ + }} {{\bf{e}}^{\ln z}} = {{\bf{e}}^{\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\ln z} \right]}} = \require{bbox} \bbox[2pt,border:1px solid black]{{{{\bf{e}}^{ - 2}}}}\]