Answer
$1$
Work Step by Step
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\lim\limits_{x \to 0^{+}} f(x)=\lim\limits_{x \to 0^{+}} \dfrac{x}{\cot (\frac{\pi}{2}-x)}$
Now, $\lim\limits_{x \to 0^{+}} \dfrac{x}{\cot (\frac{\pi}{2}-x)}=\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to 0^{+}} \dfrac{1}{\csc^2(\frac{\pi}{2}-x)}=\lim\limits_{x \to 0^{+}} \sin^2(\frac{\pi}{2}-x)$
Thus,
$\sin^2(\frac{\pi}{2}-0)=1$