1. **State the problem:** Find the left-hand limit as $x$ approaches $-2$ of the expression $$\frac{1}{x-2} - \frac{1}{x^2-4}.$$\n\n2. **Recall the formula and factorization:** Note that $x^2 - 4$ can be factored as $$(x-2)(x+2).$$\n\n3. **Rewrite the expression with common denominator:**\n$$\frac{1}{x-2} - \frac{1}{(x-2)(x+2)} = \frac{(x+2)}{(x-2)(x+2)} - \frac{1}{(x-2)(x+2)} = \frac{(x+2) - 1}{(x-2)(x+2)} = \frac{x+1}{(x-2)(x+2)}.$$\n\n4. **Evaluate the limit from the left as $x \to -2^-$.**\nSubstitute values slightly less than $-2$ to analyze the sign:\n- Numerator: $x+1$ near $-2$ is approximately $-1$ (negative).\n- Denominator: $(x-2)(x+2)$ near $-2$ is $(\text{negative} - 2)(\text{negative} + 2)$. Specifically, $(x-2)$ near $-2$ is $-4$ (negative), and $(x+2)$ approaches $0$ from the left (negative).\nSo denominator is $(\text{negative}) \times (\text{negative}) = \text{positive}$ but since $(x+2)$ approaches $0$ from the negative side, the denominator approaches $0^+$ (positive small number).\n\n5. **Determine the limit:**\nSince numerator approaches $-1$ (negative constant) and denominator approaches $0^+$ (positive very small), the fraction approaches $-\infty$.\n\n**Final answer:**\n$$\lim_{x \to -2^-} \left( \frac{1}{x-2} - \frac{1}{x^2-4} \right) = -\infty.$$