Answer
$\lim\limits_{x\to0^+}\dfrac{2}{\sin{x}}=+\infty.$
Work Step by Step
$\lim\limits_{x\to0^+}\dfrac{2}{\sin{x}}=\dfrac{2}{\sin{0^+}}=\dfrac{2}{0^+}=+\infty.$
Note: for $\theta$ (in radians) close to $0$ , $\sin{\theta}\approx\theta.$ Since, $0^+$ is extremely close to $0$, then $\sin{0^+}=0^+$
