4.10-bo'lim : L'Hospital qoidalari va noaniq shakllar
11. \(\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}}\) 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,
\[{\mbox{as}}\,\,\,x \to \infty \hspace{0.5in}{\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}} \to {\infty ^0}\]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,
\[z = {\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}}\]va ikkala tomonning jurnalini oling. Biz ham biroz soddalashtiramiz.
\[\ln z = \ln \left( {{{\left[ {{{\bf{e}}^x} + x} \right]}^{{}^{1}/{}_{x}}}} \right) = \frac{1}{x}\ln \left[ {{{\bf{e}}^x} + x} \right] = \frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x}\]3-qadam
Endi biz chegarani \(x \to \infty \) sifatida qabul qilishimiz mumkin.
\[\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x}} \right]\]Davom etishdan oldin, bizda quyidagilar borligini bilib olaylik:
\[{\mbox{as}}\,\,\,x \to \infty \hspace{0.5in}\frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x} \to = \frac{\infty }{\infty }\]va bizda L'Hospital qoidasidan foydalanishimiz mumkin bo'lgan chegaramiz bor.
4-qadam
L'Hospital qoidasini qo'llash,
\[\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x}} \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{{}^{{{{\bf{e}}^x} + 1}}/{}_{{{{\bf{e}}^x} + x}}}}{1} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x} + 1}}{{{{\bf{e}}^x} + x}}\]5-qadam
Endi bizda shunday chegara mavjud:
\[{\mbox{as}}\,\,\,x \to \infty \hspace{0.5in}\frac{{{{\bf{e}}^x} + 1}}{{{{\bf{e}}^x} + x}} \to \frac{\infty }{\infty }\]va shuning uchun biz L'Hospital qoidasidan bu borada ham foydalanishimiz mumkin. Shunday qilib,
\[\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x} + 1}}{{{{\bf{e}}^x} + x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x}}}{{{{\bf{e}}^x} + 1}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x}}}{{{{\bf{e}}^x}}} = \mathop {\lim }\limits_{x \to \infty } \left( 1 \right) = 1\]E'tibor bering, biz L'Hospital qoidasidan ikki marta foydalanishga majbur bo'ldik va biz chegarani olishimiz uchun biroz soddalashtirishga ishonch hosil qildik.
6-qadam
Endi biz shuni eslashimiz kerak,
\[z = {{\bf{e}}^{\ln z}}\]Bu, o'z navbatida, biz asl chegarani quyidagicha bajarishimiz mumkinligini anglatadi,
\[\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}} = \mathop {\lim }\limits_{x \to \infty } z = \mathop {\lim }\limits_{x \to \infty } {{\bf{e}}^{\ln z}} = {{\bf{e}}^{\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right]}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\bf{e}}}\]