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How to find the limit

$$\underset{x\to \infty }{\lim }{\left[{\int }_0^1{\left(1+\sin \frac{}{2}t\right)}^n\mathrm{d}t\right]}^{\frac{1}{n}}$$

According to the exchange rule of limit and integral, we can move the limit symbol into the integral symbol to get:

$$\begin{array}{rlrlrlr}\underset{n\to \infty }{\lim }{\left[{\int }_0^1{\left(1+\sin \frac{}{2}t\right)}^{n}dt\right]}^{\frac{1}{n}}& =\underset{n\to \infty }{\lim }{\left[{\int }_0^1{\left(1+\sin \frac{}{2}t\right)}^{n}dt\right]}^{\frac{1}{n}}& =\exp \left(\underset{n\to \infty }{\lim }\frac{\ln \left({\int }_0^1{\left(1+\sin \frac{}{2}t\right)}^{n}dt\right)}{n}\right)& =\exp \left(\underset{n\to \infty }{\lim }\frac{1}{n}\ln \left[\frac{1-\cos \frac{}{2}(n+1)}{\pi (n+1)}+\frac{2}{Pi}\right]\right)& =\exp \left(\underset{n\to \infty }{\lim }\frac{\ln \left(1-\cos \frac{}{2}(n+1)\right)}{n}+\underset{n\to \infty }{\lim }\frac{\ln \left(\frac{2}{\pi (n+1)}\right)}{n}\right)& =\exp \left(0+0\right)& =\boxed{1}.\end{array}$$

Among them, at the third equal sign, we use the integral formula ${\int }_0^1{\sin }^n(\pi /2t)dt=\frac{1-cos(\pi /2)(n+1}{\pi (n+1)}+\frac{2}{Pi}$. At the last equal sign, we take advantage of when $n\to \infty wave$ ,$\frac{\ln \left(1-\cos \frac{}{2}(n+1)\right)}{n}$和$\frac{\ln \left(\frac{2}{\pi (n+1)}\right)}{n}$The limits are all $0$。