$\text{⨍ (x)=}\frac{\sqrt{1+\mathrm{sinx}}\text{-}\sqrt{1-\mathrm{sinx}}}{\mathrm{x}}$ is continuous on $\mathrm{R}$ then $⨍ (0)=$_______

$f\left(0\right)=\underset{\mathrm{x}\to 0}{Lim} \frac{\sqrt{1+\sin \mathrm{x}}-\sqrt{1-\sin \mathrm{x}}}{\mathrm{x}}$ 

Apply L-Hospital Rule

       $=\underset{\mathrm{x}\to 0}{Lim}\frac{ {\displaystyle \frac{1}{2\sqrt{1+\sin \mathrm{x}}}}\cos \mathrm{x}-{\displaystyle \frac{1}{2\sqrt{1-\sin \mathrm{x}}}}(-\cos \mathrm{x})}{1}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=1$