Question: User: y = f(x) y = g(x) y = h(x) EX lim x\to1 f(f(x)) EX lim x\to2 h(g(x)) EX lim x\to-1 g(h(x)) Graph shapes and math: top-left: y = f(x), a piecewise graph with an open circle near $(-4, 2)$, a filled point near $(-3, -3)$, a curve descending to a filled minimum near $(1, -4)$, then rising through a filled point near $(3, 0)$ and continuing up to the right. top-center: y = g(x), a piecewise graph with a descending line from the upper left to a filled point near $(-4, 2)$, a curve starting at an open circle near $(-4, -4)$ and rising to a filled point near $(2, 3)$, then a U-shaped curve with an open circle near $(2, 1)$ continuing rightward to about $(7, 2)$. top-right: y = h(x), a piecewise graph with a descending line from about $(-5, 2)$ to an open circle near $(-1, -4)$, then a curve rising to a peak near $(2, 5)$ and falling to a filled point near $(4, 1)$.

1. **Problem:** Find $$\lim_{x \to 1} f(f(x))$$ using the given graphs. 2. **Step 1: Understand the inner limit $$\lim_{x \to 1} f(x)$$.** - From the description, at $x=1$, $f(x)$ has a filled minimum near $(1, -4)$. - So, $$\lim_{x \to 1} f(x) = -4$$. 3. **Step 2: Evaluate the outer function at this limit: $$f(-4)$$.** - From the graph of $f(x)$, near $x=-4$, there is an open circle at $(-4, 2)$, meaning $f(-4)$ is not defined at 2. - Since the circle is open, the value at $x=-4$ is not 2; the filled point near $(-3, -3)$ is irrelevant here. - The function value at $x=-4$ is undefined or different from 2. 4. **Step 3: Check if the limit $$\lim_{x \to -4} f(x)$$ exists.** - The graph shows an open circle at $(-4, 2)$ and a filled point at $(-3, -3)$. - The limit from the right of $-4$ is near $-3$, from the left is near 2. - Since the open circle is at 2, the limit at $x=-4$ is 2. 5. **Step 4: Since $f(-4)$ is undefined but the limit exists, $$\lim_{x \to -4} f(x) = 2$$.** 6. **Step 5: Therefore, $$\lim_{x \to 1} f(f(x)) = \lim_{y \to -4} f(y) = 2$$. **Final answer:** $$\boxed{2}$$