2.8-bo'lim : Infinity chegaralari, II qism
3. \(f\left( x \right) = 2{{\bf{e}}^{6x}} - {{\bf{e}}^{ - 7x}} - 10{{\bf{e}}^{4x}}\) 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 chegara uchun musbat ko'rsatkichli eksponentlar shunchaki nolga tushadi va manfiy ko'rsatkichli faqat bitta eksponensial bor (u cheksizlikka boradi) va shuning uchun bu chegara bilan bog'liq ko'p narsa yo'q.
\[\mathop {\lim }\limits_{x \to \, - \infty } \left( {2{{\bf{e}}^{6x}} - {{\bf{e}}^{ - 7x}} - 10{{\bf{e}}^{4x}}} \right) = 0 - \infty - 0 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \infty }}\]b \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\) Yechim
Bu erda bizda ijobiy 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‘alaba qozonishini” aniqlash uchun, eng katta ko‘rsatkichga ega bo‘lgan atamani ajratib ko‘rsatish va undan so‘ng asosiy chegaraviy xususiyatlardan foydalanish kerak.
\[\mathop {\lim }\limits_{x \to \,\infty } \left( {2{{\bf{e}}^{6x}} - {{\bf{e}}^{ - 7x}} - 10{{\bf{e}}^{4x}}} \right) = \mathop {\lim }\limits_{x \to \,\infty } \left[ {{{\bf{e}}^{6x}}\left( {2 - {{\bf{e}}^{ - 13x}} - 10{{\bf{e}}^{ - 2x}}} \right)} \right] = \left( \infty \right)\left( 2 \right) = \require{bbox} \bbox[2pt,border:1px solid black]{\infty }\]