Question:

Step 1: Evaluate $\lim_{x \to 0^+} g(f(x))$. As $x \to 0^+$, $f(x) = x+1$. Thus, $f(x) \to 1^+$. For $y \to 1^+$, $g(y)$ is defined by $g(y) = y^2-2y-2$ since $1 \leq y < 2$. Therefore, $$ \lim_{x \to 0^+} g(f(x)) = \lim_{y \to 1^+} g(y) = \lim_{y \to 1^+} (y^2-2y-2) = (1)^2-2(1)-2 = 1-2-2 = -3. $$ Step 2: Evaluate $\lim_{x \to 0^-} g(f(x))$. As $x \to 0^-$, $f(x) = 2-x$. Thus, $f(x) \to 2^+$. For $y \to 2^+$, $g(y)$ is defined by $g(y) = y-5$ since $y \geq 2$. Therefore, $$ \lim_{x \to 0^-} g(f(x)) = \lim_{y \to 2^+} g(y) = \lim_{y \to 2^+} (y-5) = 2-5 = -3. $$ Step 3: Evaluate $\lim_{x \to 0^+} [f(f(x))]$. As $x \to 0^+$, $f(x) = x+1$. Since $x>0$, $f(x) > 1$. Thus, $f(x) \to 1^+$. Let $y = f(x)$. As $x \to 0^+$, $y \to 1^+$. Since $y > 0$, $f(y) = y+1$. Therefore, $f(f(x)) = f(x+1) = (x+1)+1 = x+2$. As $x \to 0^+$, $x+2 \to 2^+$. For $x$ sufficiently close to $0$ and $x>0$, $x+2$ is a value slightly greater than 2. The greatest integer less than or equal to $x+2$ is $2$. Thus, $$ \lim_{x \to 0^+} [f(f(x))] = \lim_{x \to 0^+} [x+2] = 2. $$ Step 4: Evaluate $\lim_{x \to 0^-} [g(g(x))]$. As $x \to 0^-$, $g(x) = 3+x$. Since $x<0$, $3+x < 3$. Thus, $g(x) \to 3^-$. Let $y = g(x)$. As $x \to 0^-$, $y \to 3^-$. For $y$ slightly less than 3 (i.e., $y \in [2, 3)$), $g(y)$ is defined by $g(y) = y-5$. Therefore, $g(g(x)) = g(3+x) = (3+x)-5 = x-2$. As $x \to 0^-$, $x-2 \to -2^-$. For $x$ sufficiently close to $0$ and $x<0$, $x-2$ is a value slightly less than -2. The greatest integer less than or equal to $x-2$ is $-3$. Thus, $$ \lim_{x \to 0^-} [g(g(x))] = \lim_{x \to 0^-} [x-2] = -3. $$