Answer
$0$
Work Step by Step
$\lim _{x\rightarrow 0}\dfrac {1-\sqrt {1-x^{2}}}{x}=\dfrac {\left( 1-\sqrt {1-x^{2}}\right) \left( 1+\sqrt {1-x^{2}}\right) }{x\left( 1+\sqrt {1-x^{2}}\right) }=\dfrac {x^{2}}{x\left( 1+\sqrt {1-x^{2}}\right) }=\dfrac {x}{\left( 1+\sqrt {1-x^{2}}\right) }=\dfrac {0}{1+\sqrt {1}}=0$
