2.8-bo'lim : Infinity chegaralari, II qism
4. \(f\left( x \right) = 3{{\bf{e}}^{ - x}} - 8{{\bf{e}}^{ - 5x}} - {{\bf{e}}^{10x}}\) uchun quyidagi chegaralarning har birini baholang.
- \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\)
- \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\)
a \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\) Yechim
Bu erda bizda salbiy ko'rsatkichli ikkita ko'rsatkich bor va shuning uchun ikkalasi ham chegarada cheksizlikka boradi. Biroq, har bir atama qarama-qarshi belgilarga ega va shuning uchun har bir atama chegara uchun turli xil javoblarni taklif qilayotganga o'xshaydi.
Qaysi "g'olib" ekanligini aniqlash uchun biz qilishimiz kerak bo'lgan narsa - eng salbiy ko'rsatkichga ega bo'lgan atamani ajratib ko'rsatish va keyin asosiy chegara xususiyatlaridan foydalanish.
\[\mathop {\lim }\limits_{x \to \, - \infty } \left( {3{{\bf{e}}^{ - x}} - 8{{\bf{e}}^{ - 5x}} - {{\bf{e}}^{10x}}} \right) = \mathop {\lim }\limits_{x \to \, - \infty } \left[ {{{\bf{e}}^{ - 5x}}\left( {3{{\bf{e}}^{4x}} - 8 - {{\bf{e}}^{15x}}} \right)} \right] = \left( \infty \right)\left( { - 8} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \infty }}\]b \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\) Yechim
Bu chegara uchun manfiy ko'rsatkichli ko'rsatkichlar shunchaki nolga tushadi va musbat ko'rsatkichga ega bo'lgan faqat bitta eksponensial bor (u cheksizlikka boradi) va shuning uchun bu chegara bilan bog'liq ko'p narsa yo'q.
\[\mathop {\lim }\limits_{x \to \,\infty } \left( {3{{\bf{e}}^{ - x}} - 8{{\bf{e}}^{ - 5x}} - {{\bf{e}}^{10x}}} \right) = 0 - 0 - \infty = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \infty }}\]