设$f(x)$在$[0,\dfrac{1}{2}]$上连续，在$(0,\dfrac{1}{2})$内可导，且$f(0) = 0$，$f(\dfrac{1}{2})=\dfrac{3}{4}$，证明至少存在一点$\xi\in(0,\dfrac{1}{2})$，使$f'(\xi)=6\xi$.

设参数方程为$\begin{cases}x = 1+\sin2\theta\\y = 2\sin\theta + 3\end{cases}$，$\theta$为参数，求$\left.\dfrac{dy}{dx}\right|_{\theta = 0}$.

极限$\lim\limits_{x \to \infty} (1 + \dfrac{1}{x})^{3x + 1} =$______.