Question

If $f(x)=\left\{\begin{array}{ll}\sin x \neq n \pi, n \in Z \\ 0, & \text { otherwise }\end{array}\right.$ and $g(x)=\left\{\begin{array}{ll}x^2+1, & x \neq 0,2 \\ 4, & x=0 \\ 5, & x=2\end{array}\right.$, then $\lim _{x \rightarrow 0} g\{f(x)\}$ is equal to

(A) 1

(B) 0

(C) $\frac{1}{2}$

(D) $\frac{1}{4}$

Answer 1

$\lim _{x \rightarrow 0} g\{f(x)\}$

$=\lim _{x \rightarrow 0}\{f(x)\}^2+1$

$=\lim _{x \rightarrow 0} \sin ^2 x+1=1$

So, The correct option is (A).