If f(x) = begin{cases} sin x, & x ne npi, n i | vedclass

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(D) To find $\lim_{x 	o 0} g(f(x))$,we examine the left-hand and right-hand limits.For $x 	o 0$,$x 
e 0$,so $f(x) = \sin x$.As $x 	o 0$,$f(x) = \s

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$1$

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(D) To find $\lim_{x 	o 0} g(f(x))$,we examine the left-hand and right-hand limits.For $x 	o 0$,$x 
e 0$,so $f(x) = \sin x$.As $x 	o 0$,$f(x) = \sin x 	o 0$.Let $u = f(x)$. As $x 	o 0$,$u 	o 0$ through values close to $0$ but not equal to $0$ (since $\sin x 
e 0$ for $x$ near $0$ except at $x=0$).Thus,$\lim_{x 	o 0} g(f(x)) = \lim_{u 	o 0} g(u)$.Since $u 	o 0$ but $u 
e 0$,we use the definition $g(u) = u^2 + 1$.Therefore,$\lim_{u 	o 0} (u^2 + 1) = 0^2 + 1 = 1$.

If $f(x) = \begin{cases} \sin x, & x \ne n\pi, n \in \mathbb{Z} \\ 2, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \ne 0, 2 \\ 4, & x = 0 \\ 5, & x = 2 \end{cases}$,then $\lim_{x \to 0} g(f(x))$ is

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