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(B) To find $\lim _{x \rightarrow 0} g(f(x))$, we evaluate the left-hand and right-hand limits.For the left-hand limit $(x \rightarrow 0^-)$: $f(x) =

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is equal to $-5$

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(B) To find $\lim _{x \rightarrow 0} g(f(x))$, we evaluate the left-hand and right-hand limits.For the left-hand limit $(x \rightarrow 0^-)$: $f(x) = 2-x$. As $x \rightarrow 0^-$, $f(x) \rightarrow 2^+$. Since $f(x) > 2$, we use the definition $g(x) = x-7$. Thus, $\lim _{x \rightarrow 0^-} g(f(x)) = \lim _{f(x) \rightarrow 2^+} (f(x)-7) = 2-7 = -5$.For the right-hand limit $(x \rightarrow 0^+)$: $f(x) = x+2$. As $x \rightarrow 0^+$, $f(x) \rightarrow 2^+$. Again, since $f(x) > 2$, we use $g(x) = x-7$. Thus, $\lim _{x \rightarrow 0^+} g(f(x)) = \lim _{f(x) \rightarrow 2^+} (f(x)-7) = 2-7 = -5$.Since both limits are equal, $\lim _{x \rightarrow 0} g(f(x)) = -5$.

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = \begin{cases} x+2, & x>0 \\ 2-x, & x \leq 0 \end{cases}$ and $g(x) = \begin{cases} x^2-2x-2, & 1 \leq x < 2 \\ x-7, & x \geq 2 \\ x+5, & x < 1 \end{cases}$, then $\lim _{x \rightarrow 0} g(f(x))$

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