If f: R rightarrow R defined by f(x) = begin{ | vedclass

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(A) Since $f(x)$ is continuous on $R$,the left-hand limit at $x=0$ must equal $f(0)$.$f(0) = \lim_{x \rightarrow 0^-} f(x) = \lim_{h \rightarrow 0} f(

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$1/2$

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(A) Since $f(x)$ is continuous on $R$,the left-hand limit at $x=0$ must equal $f(0)$.$f(0) = \lim_{x ightarrow 0^-} f(x) = \lim_{h ightarrow 0} f(-h)$$f(0) = \lim_{h ightarrow 0} \frac{\sin(-h) - \sin(-\frac{h}{2})}{-h}$$f(0) = \lim_{h ightarrow 0} \frac{-\sin h + \sin(\frac{h}{2})}{-h}$$f(0) = \lim_{h ightarrow 0} \left[ \frac{\sin h}{h} - \frac{\sin(\frac{h}{2})}{h} ight]$$f(0) = \lim_{h ightarrow 0} \left[ \frac{\sin h}{h} - \frac{1}{2} \cdot \frac{\sin(\frac{h}{2})}{\frac{h}{2}} ight]$Using the standard limit $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$,we get:$f(0) = 1 - \frac{1}{2}(1) = \frac{1}{2}$.

$A$. $\|x\| + \|x - 2\|$	$I$. Right hand limit does not exist at $x = 2$.
$B$. $\text{cosech } x$	$II$. Continuous only for non-zero real values of $x$.
$C$. $x - [x]$	$III$. Limit is zero for all real $x$.
$D$. $\sqrt{2 - x}$	$IV$. Continuous for all real value of $x$.
	$V$. Discontinuous at all integral values of $x$.

If $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{\sin x - \sin \frac{x}{2}}{x}, & x < 0 \\ \frac{\sqrt{x^2+x} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}$ is continuous on $R$,then $f(0) = $

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