Answer
$1$
Work Step by Step
$\lim _{x\rightarrow 0}\dfrac {\arctan x}{\sin x}=\dfrac {\dfrac {d}{dx}\left( \arctan x\right) }{\dfrac {d}{dx}\left( \sin x\right) }=\dfrac {\dfrac {1}{1+x^{2}}}{\cos x}=\dfrac {\dfrac {1}{1+0}}{1}=1$
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