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本题考查数列极限的计算，核心思路是将指数形式转化为自然指数形式，利用等价无穷小替换或泰勒展开处理极限。关键在于正确展开$\sin\left(\dfrac{\pi}{4}+\dfrac{1}{n}\right)$并分析其对数的极限行为。

1. 转化指数形式
   将原式写成自然指数形式：
   $\lim _{n\rightarrow \infty }{[ \sin (\dfrac {\pi }{4}+\dfrac {1}{n})] }^{n} = \lim _{n\rightarrow \infty }{e}^{n\ln \sin \left(\dfrac {\pi }{4}+\dfrac {1}{n}\right)}.$

2. 展开$\sin\left(\dfrac{\pi}{4}+\dfrac{1}{n}\right)$
   当$n \rightarrow \infty$时，$\dfrac{1}{n} \rightarrow 0$，利用泰勒展开：
   $\sin\left(\dfrac{\pi}{4}+\dfrac{1}{n}\right) \approx \sin\dfrac{\pi}{4} + \dfrac{1}{n}\cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2n}.$

3. 近似处理对数项
   取自然对数并展开：
   $\ln\sin\left(\dfrac{\pi}{4}+\dfrac{1}{n}\right) \approx \ln\left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2n}\right) = \ln\dfrac{\sqrt{2}}{2} + \dfrac{1}{n}.$

4. 计算指数部分的极限
   代入原式并化简：
   $\lim _{n\rightarrow \infty }{e}^{n\left(\ln\dfrac{\sqrt{2}}{2} + \dfrac{1}{n}\right)} = \lim _{n\rightarrow \infty }{e}^{n\ln\dfrac{\sqrt{2}}{2} + 1}.$
   由于$\ln\dfrac{\sqrt{2}}{2} < 0$，当$n \rightarrow \infty$时，$n\ln\dfrac{\sqrt{2}}{2} \rightarrow -\infty$，因此：
   ${e}^{-\infty} = 0.$