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2.8-bo'lim : Infinity chegaralari, II qism

6. \(\displaystyle f\left( x \right) = \frac{{{{\bf{e}}^{ - 7x}} - 2{{\bf{e}}^{3x}} - {{\bf{e}}^x}}}{{{{\bf{e}}^{ - x}} + 16{{\bf{e}}^{10x}} + 2{{\bf{e}}^{ - 4x}}}}\) uchun quyidagi chegaralarning har birini baholang.

  1. \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\)
  2. \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\)

a \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\) Yechim

Salbiy ko'rsatkichli ko'rsatkichlar bu chegara uchun cheksiz bo'lgan maxrajdagi yagona atamalardir va shuning uchun biz ushbu limitni hisoblash uchun maxrajdagi eng manfiy ko'rsatkichga ega bo'lgan eksponentni (chunki u eng tez cheksizlikka boradi) suratdan ham, maxrajdan ham faktor qilishimiz kerak bo'ladi.

\[\begin{align*}\mathop {\lim }\limits_{x \to \, - \infty } \frac{{{{\bf{e}}^{ - 7x}} - 2{{\bf{e}}^{3x}} - {{\bf{e}}^x}}}{{{{\bf{e}}^{ - x}} + 16{{\bf{e}}^{10x}} + 2{{\bf{e}}^{ - 4x}}}} & = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{{\bf{e}}^{ - 4x}}\left( {{{\bf{e}}^{ - 3x}} - 2{{\bf{e}}^{7x}} - {{\bf{e}}^{5x}}} \right)}}{{{{\bf{e}}^{ - 4x}}\left( {{{\bf{e}}^{3x}} + 16{{\bf{e}}^{14x}} + 2} \right)}}\\ & = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{{\bf{e}}^{ - 3x}} - 2{{\bf{e}}^{7x}} - {{\bf{e}}^{5x}}}}{{{{\bf{e}}^{3x}} + 16{{\bf{e}}^{14x}} + 2}} = \frac{{\infty - 0 - 0}}{{0 + 0 + 2}} = \require{bbox} \bbox[2pt,border:1px solid black]{\infty }\end{align*}\]

b \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\) Yechim

Ijobiy ko'rsatkichli ko'rsatkichlar bu chegara uchun cheksiz bo'lgan maxrajdagi yagona atamalardir va shuning uchun biz ushbu limitni hisoblash uchun maxrajdagi eng musbat ko'rsatkichga ega bo'lgan ko'rsatkichni (chunki u eng tez cheksizlikka boradi) suratdan ham, maxrajdan ham faktor qilishimiz kerak.

\[\begin{align*}\mathop {\lim }\limits_{x \to \, \infty } \frac{{{{\bf{e}}^{ - 7x}} - 2{{\bf{e}}^{3x}} - {{\bf{e}}^x}}}{{{{\bf{e}}^{ - x}} + 16{{\bf{e}}^{10x}} + 2{{\bf{e}}^{ - 4x}}}} & = \mathop {\lim }\limits_{x \to \, \infty } \frac{{{{\bf{e}}^{10x}}\left( {{{\bf{e}}^{ - 17x}} - 2{{\bf{e}}^{ - 7x}} - {{\bf{e}}^{ - 9x}}} \right)}}{{{{\bf{e}}^{10x}}\left( {{{\bf{e}}^{ - 11x}} + 16 + 2{{\bf{e}}^{ - 14x}}} \right)}}\\ & = \mathop {\lim }\limits_{x \to \, \infty } \frac{{{{\bf{e}}^{ - 17x}} - 2{{\bf{e}}^{ - 7x}} - {{\bf{e}}^{ - 9x}}}}{{{{\bf{e}}^{ - 11x}} + 16 + 2{{\bf{e}}^{ - 14x}}}} = \frac{{0 - 0 - 0}}{{0 + 16 + 0}} = \require{bbox} \bbox[2pt,border:1px solid black]{0}\end{align*}\]