The exponentials with the negative exponents are the only terms in the denominator going to infinity for this limit and so we’ll need to factor the exponential with the most negative exponent in the denominator (because it will be going to infinity fastest) from both the numerator and denominator to evaluate this limit.

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

The exponentials with the positive exponents are the only terms in the denominator going to infinity for this limit and so we’ll need to factor the exponential with the most positive exponent in the denominator (because it will be going to infinity fastest) from both the numerator and denominator to evaluate this limit.

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