1. The problem is to find the limit $$\lim_{x \to -\infty} \tanh(x) = \lim_{x \to -\infty} \frac{e^x - e^{-x}}{e^x + e^{-x}}$$ and understand why the sandwich theorem was not used. 2. Recall the definition of hyperbolic tangent: $$\tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$. 3. When evaluating limits involving exponentials as $x \to -\infty$, note that $e^x \to 0$ and $e^{-x} \to \infty$. 4. Direct substitution gives an indeterminate form $$\frac{0 - \infty}{0 + \infty} = \frac{-\infty}{\infty}$$ which is not immediately solvable. 5. To resolve this, divide numerator and denominator by the dominant term in the denominator, which is $e^{-x}$ (since it tends to $+\infty$): $$\lim_{x \to -\infty} \frac{e^x - e^{-x}}{e^x + e^{-x}} = \lim_{x \to -\infty} \frac{\frac{e^x}{e^{-x}} - \frac{e^{-x}}{e^{-x}}}{\frac{e^x}{e^{-x}} + \frac{e^{-x}}{e^{-x}}} = \lim_{x \to -\infty} \frac{e^{2x} - 1}{e^{2x} + 1}$$ 6. As $x \to -\infty$, $e^{2x} \to 0$, so the limit becomes: $$\frac{0 - 1}{0 + 1} = \frac{-1}{1} = -1$$ 7. The sandwich theorem was not used because the function $\tanh(x)$ is continuous and the limit can be evaluated directly by algebraic manipulation and understanding the behavior of exponentials at infinity. 8. The sandwich theorem is typically used when a function is bounded between two functions whose limits are known and equal. Here, the direct algebraic approach is simpler and sufficient. Final answer: $$\lim_{x \to -\infty} \tanh(x) = -1$$