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If $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ defined by
$f(x)=\left\{\begin{array}{ll}\frac{\sin x-\sin \frac{x}{2}}{x}, & x < 0 \\ \frac{\sqrt{x^2+x}-\sqrt{x}}{x^{3 / 2}}, & x>0\end{array}\right.$ is continuous on $\mathbb{R}$, then $f(0)=$
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$f(x)=\left\{\begin{array}{ll}\frac{\sin x-\sin \frac{x}{2}}{x}, & x < 0 \\ \frac{\sqrt{x^2+x}-\sqrt{x}}{x^{3 / 2}}, & x>0\end{array}\right.$ is continuous on $\mathbb{R}$, then $f(0)=$
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The correct answer is:
$1 / 2$
Since $f(x)$ is also continuous on $\mathbf{R}$
$$
\begin{aligned}
& \text { Hence } f(0)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} \frac{\sin (-h)-\sin \left(-\frac{h}{2}\right)}{(-h)} \\
& \Rightarrow f(0)=\lim _{h \rightarrow 0}\left[\frac{\sin h}{h}-\frac{1}{2} \cdot \frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right]=1-\frac{1}{2} \\
& \Rightarrow f(0)=\frac{1}{2}
\end{aligned}
$$
$$
\begin{aligned}
& \text { Hence } f(0)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} \frac{\sin (-h)-\sin \left(-\frac{h}{2}\right)}{(-h)} \\
& \Rightarrow f(0)=\lim _{h \rightarrow 0}\left[\frac{\sin h}{h}-\frac{1}{2} \cdot \frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right]=1-\frac{1}{2} \\
& \Rightarrow f(0)=\frac{1}{2}
\end{aligned}
$$
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