The exponential with the negative exponent is the only term in the denominator going to infinity for this limit and so we’ll need to factor the exponential with the negative exponent in the denominator from both the numerator and denominator to evaluate this limit.

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

The exponential with the positive exponent is the only term in the denominator going to infinity for this limit and so we’ll need to factor the exponential with the positive exponent in the denominator from both the numerator and denominator to evaluate this limit.

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