If $f(x)=\left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}$ is continuous at $x=0$, then $f(0)=$

The function $f(x)=|x-24|$ is

$$\mathop {\lim }\limits_{n \to \infty }\left(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2-1}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right)= $$

$$\mathop {\lim }\limits_{x \to 0} \left( {{{\sin (\pi {{\cos }^2}x} \over {{x^2}}}} \right) = $$