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If $f(x)=x-[x], x \in R$, then $f^{\prime}\left(\frac{1}{2}\right)$ is equal to
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Since, $f(x)=x-[x]$
$$
L f^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{f\left(\frac{1}{2}-h\right)-f\left(\frac{1}{2}\right)}{-h}
$$
$=\lim _{h \rightarrow 0} \frac{\left(\frac{1}{2}-h\right)-\left[\frac{1}{2}-h\right]-\frac{1}{2}+\left[\frac{1}{2}\right]}{-h}$
$$
=\lim _{h \rightarrow 0} \frac{\frac{1}{2}-h-0-\frac{1}{2}+0}{-h}=1
$$
and $R f^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{f\left(\frac{1}{2}+h\right)-f\left(\frac{1}{2}\right)}{h}$
$$
=\lim _{h \rightarrow 0} \frac{\left(\frac{1}{2}+h\right)-\left[\frac{1}{2}+h\right]-\frac{1}{2}+\left[\frac{1}{2}\right]}{h}
$$
$$
=\lim _{h \rightarrow 0} \frac{\frac{1}{2}+h-0-\frac{1}{2}+0}{h}=1
$$
As, $L f^{\prime}\left(\frac{1}{2}\right)=R f^{\prime}\left(\frac{1}{2}\right)$. Hence, $f^{\prime}\left(\frac{1}{2}\right)=1$
$$
L f^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{f\left(\frac{1}{2}-h\right)-f\left(\frac{1}{2}\right)}{-h}
$$
$=\lim _{h \rightarrow 0} \frac{\left(\frac{1}{2}-h\right)-\left[\frac{1}{2}-h\right]-\frac{1}{2}+\left[\frac{1}{2}\right]}{-h}$
$$
=\lim _{h \rightarrow 0} \frac{\frac{1}{2}-h-0-\frac{1}{2}+0}{-h}=1
$$
and $R f^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{f\left(\frac{1}{2}+h\right)-f\left(\frac{1}{2}\right)}{h}$
$$
=\lim _{h \rightarrow 0} \frac{\left(\frac{1}{2}+h\right)-\left[\frac{1}{2}+h\right]-\frac{1}{2}+\left[\frac{1}{2}\right]}{h}
$$
$$
=\lim _{h \rightarrow 0} \frac{\frac{1}{2}+h-0-\frac{1}{2}+0}{h}=1
$$
As, $L f^{\prime}\left(\frac{1}{2}\right)=R f^{\prime}\left(\frac{1}{2}\right)$. Hence, $f^{\prime}\left(\frac{1}{2}\right)=1$
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