1. **State the problem:** Find the limit $$\lim_{x \to -1} 9 \cdot g(x)$$ where $$g(x) = \begin{cases} 5 - 2x & \text{if } x < -1 \\ 2x & \text{if } x \geq -1 \end{cases}$$. 2. **Understand the piecewise function:** Since the limit is as $$x$$ approaches $$-1$$, we need to check the left-hand limit (as $$x \to -1^-$$) and the right-hand limit (as $$x \to -1^+$$) of $$g(x)$$. 3. **Calculate left-hand limit of $$g(x)$$:** For $$x < -1$$, $$g(x) = 5 - 2x$$. $$\lim_{x \to -1^-} g(x) = 5 - 2(-1) = 5 + 2 = 7$$. 4. **Calculate right-hand limit of $$g(x)$$:** For $$x \geq -1$$, $$g(x) = 2x$$. $$\lim_{x \to -1^+} g(x) = 2(-1) = -2$$. 5. **Check if the limit of $$g(x)$$ exists at $$x = -1$$:** Since left-hand limit $$7 \neq -2$$ right-hand limit, the limit $$\lim_{x \to -1} g(x)$$ does not exist. 6. **Calculate the limit of $$9 \cdot g(x)$$:** Because $$\lim_{x \to -1} g(x)$$ does not exist, $$\lim_{x \to -1} 9 \cdot g(x)$$ also does not exist. **Final answer:** $$\lim_{x \to -1} 9 \cdot g(x)$$ does not exist because the left and right limits of $$g(x)$$ at $$x = -1$$ are different.