Salbiy ko'rsatkichli eksponensial bu chegara uchun cheksiz bo'lgan maxrajdagi yagona atamadir va shuning uchun biz ushbu limitni hisoblash uchun maxrajdagi manfiy ko'rsatkichli eksponensialni ham, ham maxrajdan ko'paytirishimiz kerak bo'ladi.

\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{{{\bf{e}}^{ - 3x}} - 2{{\bf{e}}^{8x}}}}{{9{{\bf{e}}^{8x}} - 7{{\bf{e}}^{ - 3x}}}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{{\bf{e}}^{ - 3x}}\left( {1 - 2{{\bf{e}}^{11x}}} \right)}}{{{{\bf{e}}^{ - 3x}}\left( {9{{\bf{e}}^{11x}} - 7} \right)}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{1 - 2{{\bf{e}}^{11x}}}}{{9{{\bf{e}}^{11x}} - 7}} = \frac{{1 - 0}}{{0 - 7}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{7}}}\]

Ijobiy ko'rsatkichli ko'rsatkich bu chegara uchun cheksiz bo'lgan maxrajdagi yagona atamadir va shuning uchun biz bu limitni hisoblash uchun maxrajdagi musbat ko'rsatkich bilan ko'rsatkichni ham, ham maxrajdan ko'paytirishimiz kerak.

\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{{{\bf{e}}^{ - 3x}} - 2{{\bf{e}}^{8x}}}}{{9{{\bf{e}}^{8x}} - 7{{\bf{e}}^{ - 3x}}}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{{{\bf{e}}^{8x}}\left( {{{\bf{e}}^{ - 11x}} - 2} \right)}}{{{{\bf{e}}^{8x}}\left( {9 - 7{{\bf{e}}^{ - 11x}}} \right)}} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{{{\bf{e}}^{ - 11x}} - 2}}{{9 - 7{{\bf{e}}^{ - 11x}}}} = \frac{{0 - 2}}{{9 - 0}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{2}{9}}}\]