如何求极限

lim⁡x→∞[∫01(1+sin⁡π2t)ndt]1n\lim_{x \to \infty} \left [ \int_{0}^{1} \left ( 1+ \sin \frac{\pi }{2}t \right )^{n} \mathrm{d}t \right ] ^ \frac{1}{n} x→∞lim​[∫01​(1+sin2π​t)ndt]n1​

根据极限与积分的交换规则，我们可以将极限符号移入积分符号中，得到：

lim⁡n→∞[∫01(1+sin⁡π2t)ndt]1n=lim⁡n→∞[∫01(1+sin⁡π2t)ndt]1n=exp⁡(lim⁡n→∞ln⁡(∫01(1+sin⁡π2t)ndt)n)=exp⁡(lim⁡n→∞1nln⁡[1−cos⁡π2(n+1)π(n+1)+2π])=exp⁡(lim⁡n→∞ln⁡(1−cos⁡π2(n+1))n+lim⁡n→∞ln⁡(2π(n+1))n)=exp⁡(0+0)=1.\begin{aligned} \lim_{n\to \infty}\left[\int_{0}^{1}\left(1+\sin\frac{\pi}{2}t\right)^n dt\right]^\frac{1}{n}&=\lim_{n\to\infty}\left[\int_{0}^{1}\left(1+\sin\frac{\pi}{2}t\right)^n dt\right]^{\frac{1}{n}}\ &=\exp\left(\lim_{n\to\infty}\frac{\ln\left(\int_{0}^{1}\left(1+\sin\frac{\pi}{2}t\right)^n dt\right)}{n}\right)\ &=\exp\left(\lim_{n\to\infty}\frac{1}{n}\ln\left[\frac{1-\cos\frac{\pi}{2}(n+1)}{\pi(n+1)}+\frac{2}{\pi}\right]\right)\ &=\exp\left(\lim_{n\to\infty}\frac{\ln\left(1-\cos\frac{\pi}{2}(n+1)\right)}{n}+\lim_{n\to\infty}\frac{\ln\left(\frac{2}{\pi(n+1)}\right)}{n}\right)\ &=\exp\left(0+0\right)\ &=\boxed{1}. \end{aligned}n→∞lim​[∫01​(1+sin2π​t)ndt]n1​​=n→∞lim​[∫01​(1+sin2π​t)ndt]n1​ ​=exp​n→∞lim​nln(∫01​(1+sin2π​t)ndt)​​ ​=exp(n→∞lim​n1​ln[π(n+1)1−cos2π​(n+1)​+π2​]) ​=exp​n→∞lim​nln(1−cos2π​(n+1))​+n→∞lim​nln(π(n+1)2​)​​ ​=exp(0+0) ​=1​.​

其中，在第三个等号处，我们使用了积分公式∫01sin⁡n(π/2t)dt=1−cos⁡(π/2)(n+1)π(n+1)+2π\int_0^1 \sin^{n} (\pi/2 t) dt= \frac{1-\cos(\pi/2)(n+1)}{\pi(n+1)}+\frac{2}{\pi}∫01​sinn(π/2t)dt=π(n+1)1−cos(π/2)(n+1)​+π2​。在最后一个等号处，我们利用了当n→∞n\to \inftyn→∞时，ln⁡(1−cos⁡π2(n+1))n\frac{\ln\left(1-\cos\frac{\pi}{2}(n+1)\right)}{n}nln(1−cos2π​(n+1))​和ln⁡(2π(n+1))n\frac{\ln\left(\frac{2}{\pi(n+1)}\right)}{n}nln(π(n+1)2​)​的极限均为000。