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(A) To find the graph of the function $f(x) = \lim_{n 	o \infty} \frac{2}{\pi} 	an^{-1}(nx)$,we analyze the limit for different values of $x$:
$1$.

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(A) To find the graph of the function $f(x) = \lim_{n o \infty} \frac{2}{\pi} an^{-1}(nx)$,we analyze the limit for different values of $x$: $1$. If $x > 0$,then as $n o \infty$,$nx o \infty$. Therefore,$ an^{-1}(nx) o \frac{\pi}{2}$. Thus,$f(x) = \frac{2}{\pi} imes \frac{\pi}{2} = 1$. $2$. If $x = 0$,then $nx = 0$ for all $n$. Therefore,$ an^{-1}(0) = 0$. Thus,$f(x) = \frac{2}{\pi} imes 0 = 0$. $3$. If $x < 0$,then as $n o \infty$,$nx o -\infty$. Therefore,$ an^{-1}(nx) o -\frac{\pi}{2}$. Thus,$f(x) = \frac{2}{\pi} imes (-\frac{\pi}{2}) = -1$. Combining these,the function is defined as: $f(x) = \begin{cases} 1, & x > 0 \ 0, & x = 0 \ -1, & x < 0 \end{cases}$ This corresponds to the signum function,which is represented by the graph where $y=1$ for $x>0$,$y=0$ at $x=0$,and $y=-1$ for $x < 0$. This matches the graph in option $A$.

Which one of the following best represents the graph of the function $f(x) = \lim_{n \to \infty} \frac{2}{\pi} \tan^{-1}(nx)$?

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