1. **Problem statement:** Find the limit $$\lim_{x \to -\frac{1}{2}^-} \frac{x^7 + ax + \frac{1}{x^7}}{x - \frac{1}{x}} = b$$ where $a$ and $b$ are constants. 2. **Rewrite the expression:** The limit is $$\lim_{x \to -\frac{1}{2}^-} \frac{x^7 + ax + x^{-7}}{x - x^{-1}}.$$ 3. **Analyze the denominator:** $$x - \frac{1}{x} = \frac{x^2 - 1}{x}.$$ At $x = -\frac{1}{2}$, $$x^2 - 1 = \left(-\frac{1}{2}\right)^2 - 1 = \frac{1}{4} - 1 = -\frac{3}{4} \neq 0,$$ so the denominator is not zero and the limit is finite if the numerator is finite. 4. **Evaluate numerator at $x = -\frac{1}{2}$:** $$x^7 = \left(-\frac{1}{2}\right)^7 = -\frac{1}{128},$$ $$x^{-7} = \left(-\frac{1}{2}\right)^{-7} = \left(-2\right)^7 = -128,$$ so numerator is $$-\frac{1}{128} + a \left(-\frac{1}{2}\right) - 128 = -\frac{1}{128} - \frac{a}{2} - 128.$$ 5. **Evaluate denominator at $x = -\frac{1}{2}$:** $$x - \frac{1}{x} = -\frac{1}{2} - \left(-2\right) = -\frac{1}{2} + 2 = \frac{3}{2}.$$ 6. **Limit value:** $$b = \frac{-\frac{1}{128} - \frac{a}{2} - 128}{\frac{3}{2}} = \frac{-\frac{1}{128} - \frac{a}{2} - 128}{1.5}.$$ 7. **Given options for $b$ are 0, 1, -1, and 1/2.** We solve for $a$ for each: - For $b=0$: $$0 = \frac{-\frac{1}{128} - \frac{a}{2} - 128}{\frac{3}{2}} \implies -\frac{1}{128} - \frac{a}{2} - 128 = 0,$$ $$-\frac{a}{2} = 128 + \frac{1}{128},$$ $$a = -2 \left(128 + \frac{1}{128}\right) = -256 - \frac{2}{128} = -256 - \frac{1}{64} = -256.015625.$$ - For $b=1$: $$1 = \frac{-\frac{1}{128} - \frac{a}{2} - 128}{\frac{3}{2}} \implies -\frac{1}{128} - \frac{a}{2} - 128 = \frac{3}{2},$$ $$-\frac{a}{2} = 128 + \frac{1}{128} + \frac{3}{2} = 128 + 0.0078125 + 1.5 = 129.5078125,$$ $$a = -2 \times 129.5078125 = -259.015625.$$ - For $b=-1$: $$-1 = \frac{-\frac{1}{128} - \frac{a}{2} - 128}{\frac{3}{2}} \implies -\frac{1}{128} - \frac{a}{2} - 128 = -\frac{3}{2},$$ $$-\frac{a}{2} = 128 + \frac{1}{128} - \frac{3}{2} = 128 + 0.0078125 - 1.5 = 126.5078125,$$ $$a = -2 \times 126.5078125 = -253.015625.$$ - For $b=\frac{1}{2}$: $$\frac{1}{2} = \frac{-\frac{1}{128} - \frac{a}{2} - 128}{\frac{3}{2}} \implies -\frac{1}{128} - \frac{a}{2} - 128 = \frac{3}{4},$$ $$-\frac{a}{2} = 128 + \frac{1}{128} + \frac{3}{4} = 128 + 0.0078125 + 0.75 = 128.7578125,$$ $$a = -2 \times 128.7578125 = -257.515625.$$ **Summary:** The limit exists and equals $b$ for corresponding $a$ values above. **Final answer:** The limit equals $b$ as given, with $a$ values depending on $b$.