For the function f(x) = lim_{n to infty} frac | vedclass

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Question

(C) Case $1$: If $x \in \mathbb{Z}$,then $\sin(\pi x) = 0$. Thus,$f(x) = \lim_{n 	o \infty} \frac{1}{1 + n(0)} = 1$. Case $2$: If $x 
otin \mathbb{Z

Vedclass · Accepted Answer

$f$ is discontinuous for all $x \in \mathbb{Z}$

Vedclass · Answer

(C) Case $1$: If $x \in \mathbb{Z}$,then $\sin(\pi x) = 0$. Thus,$f(x) = \lim_{n 	o \infty} \frac{1}{1 + n(0)} = 1$. Case $2$: If $x 
otin \mathbb{Z}$,then $\sin^2(\pi x) > 0$. As $n 	o \infty$,$n \sin^2(\pi x) 	o \infty$. Thus,$f(x) = \lim_{n 	o \infty} \frac{1}{1 + n \sin^2(\pi x)} = 0$. Therefore,$f(x) = \begin{cases} 1, & x \in \mathbb{Z} \ 0, & x 
otin \mathbb{Z} \end{cases}$. Since the function jumps between $0$ and $1$ at every integer value,$f$ is discontinuous for all $x \in \mathbb{Z}$.

For the function $f(x) = \lim_{n \to \infty} \frac{1}{1 + n \sin^2(\pi x)}$,which of the following holds?

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