Evaluate: \[ \lim _{x \rightarrow-\infty} \frac{e^{-7 x}-2 e^{3 x}-e^{x}}{e^{-x}+16 e^{-10 x}+2 e^{-4 x}} \]

To evaluate the limit \[ \lim _{x \rightarrow-\infty} \frac{e^{-7 x}-2 e^{3 x}-e^{x}}{e^{-x}+16 e^{-10 x}+2 e^{-4 x}}, \] we begin by analyzing the behavior of each term as \( x \to -\infty \). ### Numerator Analysis: 1. **Term \( e^{-7x} \)**: As \( x \to -\infty \), \( -7x \to +\infty \), so \( e^{-7x} \to +\infty \). 2. **Term \( -2e^{3x} \)**: As \( x \to -\infty \), \( e^{3x} \to 0 \), thus \( -2e^{3x} \to 0 \). 3. **Term \( -e^x \)**: Similarly, as \( x \to -\infty \), \( e^x \to 0 \), so \( -e^x \to 0 \). Thus, the dominant term in the numerator is \( e^{-7x} \to +\infty \). ### Denominator Analysis: 1. **Term \( e^{-x} \)**: As \( x \to -\infty \), \( -x \to +\infty \), hence \( e^{-x} \to +\infty \). 2. **Term \( 16e^{-10x} \)**: Here, \( -10x \to +\infty \), therefore \( e^{-10x} \to +\infty \) leading to \( 16e^{-10x} \to +\infty \). 3. **Term \( 2e^{-4x} \)**: Similarly, \( -4x \to +\infty \) implies \( e^{-4x} \to +\infty \), thus \( 2e^{-4x} \to +\infty \). Among these terms, we need to identify which grows the fastest as \( x \to -\infty \). Since exponential functions grow more quickly with increasing coefficients of their inputs, \( 16e^{-10x} \to +\infty \) grows fastest because of the highest coefficient of \( -10x \). ### Limit Calculation: Now, we rewrite the limit with the identified dominant terms. Both the numerator and denominator tend to \( +\infty \). We can simplify by factoring out the dominating exponentials: 1. In the **numerator**, factor out \( e^{-7x} \): \[ e^{-7x}(1 - 2 e^{10x} - e^{-6x}). \] Here, as \( x \to -\infty \), \( e^{10x} \to 0 \) and \( e^{-6x} \to +\infty\). We still have the overall behavior dominated by \( e^{-7x} \). 2. In the **denominator**, factor out \( e^{-10x} \): \[ e^{-10x}(16 + 2 e^{6x} + e^{4x}). \] Again, as \( x \to -\infty \), the term \( e^{6x} \) and \( e^{4x} \) tend towards 0. Now substituting back into the limit gives us: \[ \lim_{x \to -\infty} \frac{e^{-7x}(1 - 2e^{10x} - e^{-6x})}{e^{-10x}(16 + 2e^{6x} + e^{4x})} = \lim_{x \to -\infty} \frac{e^{3x}(1 - 2e^{10x} - e^{-6x})}{16 + 2e^{6x} + e^{4x}}. \] As \( x \to -\infty\), \( e^{3x} \to 0\). Therefore, both \( 1 - 2e^{10x} \to 1 \) and \( 16 + 2e^{6x} + e^{4x} \to 16\), leading to: \[ = \frac{0}{16} = 0. \] Finally, the limit is \[ \boxed{0}. \]