To solve $\lim_{x \to 2^-} [f(x) + g(x)]$, we analyze the left-hand limits of the given functions $f(x)$ (red graph) and $g(x)$ (blue graph) as $x$ approaches $2$ from the left.

1. For $f(x)$ as $x \to 2^-$:

   • Observe the red graph. As $x$ approaches $2$ from the left, the $y$-coordinate (function value) of $f(x)$ approaches $1$.
   • Thus, $\lim_{x \to 2^-} f(x) = 1$.

2. For $g(x)$ as $x \to 2^-$:

   • Observe the blue graph. As $x$ approaches $2$ from the left, the $y$-coordinate of $g(x)$ approaches $1$.
   • Thus, $\lim_{x \to 2^-} g(x) = 1$.

3. Combine the limits: $\lim_{x \to 2^-} [f(x) + g(x)] = \lim_{x \to 2^-} f(x) + \lim_{x \to 2^-} g(x) = 1 + 1 = 2.$

Final Answer:

$\lim_{x \to 2^-} [f(x) + g(x)] = 2.$

Let me know if you'd like further clarification or additional steps!

Related Questions:

1. What is $\lim_{x \to 2^+} [f(x) + g(x)]$?
2. Analyze the continuity of $f(x)$ and $g(x)$ at $x = 2$.
3. Find $\lim_{x \to 1^-} [f(x) - g(x)]$.
4. Explore the behavior of $f(x)$ and $g(x)$ at $x = 3$.
5. What is the significance of a one-sided limit in the context of these graphs?

Tip: Always examine the graph closely for jump discontinuities or holes when calculating one-sided limits.