1. **State the problem:** Find the value of the limit $$\lim_{x \to +\infty} \frac{2^{-x}}{2^x}$$. 2. **Recall the properties of exponents:** - For any base $a > 0$, $a^{-x} = \frac{1}{a^x}$. - When dividing powers with the same base, subtract the exponents: $$\frac{a^m}{a^n} = a^{m-n}$$. 3. **Apply these properties to the expression:** $$\frac{2^{-x}}{2^x} = 2^{-x - x} = 2^{-2x}$$. 4. **Evaluate the limit:** As $x \to +\infty$, $-2x \to -\infty$, so $$2^{-2x} = \frac{1}{2^{2x}} \to 0$$ because the denominator grows without bound. 5. **Conclusion:** The limit is $$0$$. **Final answer:** c. 0