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