Answer
$1$
Work Step by Step
$\lim\limits_{x \to 0^{+}} f(x)=\lim\limits_{x \to 0^{+}} \dfrac{x}{\cot (\frac{\pi}{2}-x)}$
But $\lim\limits_{x \to 0^{+}} \dfrac{x}{\cot (\frac{\pi}{2}-x)}=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 0^{+}} \dfrac{1}{\csc^2(\frac{\pi}{2}-x)}=\lim\limits_{x \to 0^{+}} [\sin^2(\frac{\pi}{2}-x)]$
and $\sin^2(\dfrac{\pi}{2}-0)=1$
