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

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(D) We need to evaluate $\lim_{x ightarrow 0} g(f(x))$.As $x ightarrow 0$,$x 
eq n\pi$ for $n 
eq 0$. Thus,$f(x) = \sin x$.As $x ightarrow 0$,

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$1$

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(D) We need to evaluate $\lim_{x ightarrow 0} g(f(x))$.As $x ightarrow 0$,$x 
eq n\pi$ for $n 
eq 0$. Thus,$f(x) = \sin x$.As $x ightarrow 0$,$f(x) = \sin x ightarrow 0$.Since $f(x) ightarrow 0$ but $f(x) 
eq 0$ for $x$ in a deleted neighborhood of $0$,we look at the definition of $g(u)$ as $u ightarrow 0$.For $u 
eq 0, 2$,$g(u) = u^2 + 1$.Therefore,$\lim_{u ightarrow 0} g(u) = \lim_{u ightarrow 0} (u^2 + 1) = 0^2 + 1 = 1$.Thus,$\lim_{x ightarrow 0} g(f(x)) = 1$.

If $f(x) = \begin{cases} \sin x, & x \neq n\pi, n \in I \\ 2, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \neq 0, 2 \\ 2, & x = 0 \\ 4, & x = 2 \end{cases}$,then find $\lim_{x \rightarrow 0} g(f(x))$.

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