4.10-bo'lim : L'Hospital qoidalari va noaniq shakllar
7. \(\mathop {\lim }\limits_{t \to \infty } \left[ {t\ln \left( {1 + \displaystyle \frac{3}{t}} \right)} \right]\) 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 ma'lum bir ratsional funktsiyalar sinfida ishlaydi va bu ratsional funktsiya emasligi aniq.
Ammo e'tibor bering, u quyidagi noaniq shaklda,
\[{\mbox{as}}\,\,\,t \to \infty \hspace{0.5in}t\ln \left( {1 + \frac{3}{t}} \right) \to \left( \infty \right)\left( 0 \right)\]va biz ushbu bo'lim uchun eslatmalarda muhokama qilganimizdek, biz har doim bunday noaniq shaklni L'Hospital qoidasini qo'llash imkonini beradigan oqilona ifodaga aylantirishimiz mumkin.
2-qadam
Haqiqiy savol shundaki, biz birinchi atamani yoki ikkinchi atamani maxrajga o'tkazamizmi? Ko'rinishidan, birinchi atamani maxrajga ko'chirish yaxshi bo'ladi.
\[\mathop {\lim }\limits_{t \to \infty } \left[ {t\ln \left( {1 + \frac{3}{t}} \right)} \right] = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln \left( {1 + \frac{3}{t}} \right)}}{{{}^{1}/{}_{t}}}\]Shuni ham e'tiborga oling,
\[{\mbox{as}}\,\,\,t \to \infty \hspace{0.5in}\frac{{\ln \left( {1 + \frac{3}{t}} \right)}}{{{}^{1}/{}_{t}}} \to \frac{0}{0}\]va biz bu borada L'Hospital qoidasidan foydalanishimiz mumkin.
3-qadam
L'Hospital qoidasini qo'llash,
\[\mathop {\lim }\limits_{t \to \infty } \left[ {t\ln \left( {1 + \frac{3}{t}} \right)} \right] = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln \left( {1 + \frac{3}{t}} \right)}}{{{}^{1}/{}_{t}}} = \mathop {\lim }\limits_{t \to \infty } \frac{{\frac{{ - {}^{3}/{}_{{{t^2}}}}}{{1 + {}^{3}/{}_{t}}}}}{{ - {}^{1}/{}_{{{t^2}}}}}\]Nima uchun biz \(t\) ni denominatorga ko'chirishni tanlaganimizni tushuna olasizmi? Logarifmni siljitish bizni juda chalkash hosila qoldirishi mumkin edi! Bu biz uchun yaxshi ishlagan bo'lishi mumkin edi, lekin ish sezilarli darajada oshadi.
4-qadam
Biz lotinni olganimizdan keyin soddalashtirishni unutmang. Agar shunday qilsak, bu muammo juda oddiy bo'ladi.
\[\mathop {\lim }\limits_{t \to \infty } \left[ {t\ln \left( {1 + \frac{3}{t}} \right)} \right] = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln \left( {1 + \frac{3}{t}} \right)}}{{{}^{1}/{}_{t}}} = \mathop {\lim }\limits_{t \to \infty } \frac{3}{{1 + {}^{3}/{}_{t}}} = \require{bbox} \bbox[2pt,border:1px solid black]{3}\]