The graph of the function f(x) = x + frac{1}{ | vedclass

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(B) Given $f(x) = x + \frac{1}{8} \sin(2 \pi x)$ for $x \in [0, 1]$.From the graph,we observe that $f(x) > x$ for $x \in (0, 1/2)$,$f(x) < x$ for $x \

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$II$ only

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(B) Given $f(x) = x + \frac{1}{8} \sin(2 \pi x)$ for $x \in [0, 1]$.From the graph,we observe that $f(x) > x$ for $x \in (0, 1/2)$,$f(x) For $x \in (0, 1/2)$,the sequence $f_n(x)$ is strictly increasing and bounded above by $1/2$,so $\lim_{n \rightarrow \infty} f_n(x) = 1/2$.For $x \in (1/2, 1)$,the sequence $f_n(x)$ is strictly decreasing and bounded below by $1/2$,so $\lim_{n \rightarrow \infty} f_n(x) = 1/2$.For $x = 0$,$f_n(0) = 0$ for all $n$,so the limit is $0$.For $x = 1$,$f_n(1) = 1$ for all $n$,so the limit is $1$.For $x = 1/2$,$f_n(1/2) = 1/2$ for all $n$,so the limit is $1/2$.Thus,the limit is $1/2$ for all $x \in (0, 1)$. Since there are infinitely many such $x$,statement $II$ is true.Statements $I$ and $III$ are false because the limit is $0$ only at $x=0$ and $1$ only at $x=1$.Statement $IV$ is false because the limit exists for all $x \in [0, 1]$.Therefore,only statement $II$ is true.

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