Answer
$\dfrac {5}{9}$
Work Step by Step
$\lim _{x\rightarrow 0}\dfrac {\sin 5x}{\tan 9x}=\dfrac {\dfrac {d}{dx}\left( \sin 5x\right) }{\dfrac {d}{dx}\left( \tan 9x\right) }=\dfrac {5\cos 5x}{\dfrac {1}{\cos ^{2}9x}\times 9}=\dfrac {5\times \cos 0}{9\times \cos 0}=\dfrac {5}{9}$
