大一/高数/好题

$$\lim_{x\to\frac\pi2}(\sin x)^{\tan x}$$ $$\begin{aligned}\lim_{x\to\frac\pi2}(\sin x)^{\tan x} &=\lim_{x\to\frac\pi2}e^{\tan x\ln\sin x}\\&=\lim_{t\to1^-} e^{\tan x\ln t}\end{aligned}$$

法1

$$\begin{aligned}\because\lim_{t\to1^-} {\tan x\ln t}&=\lim_{t\to1^-}\frac{t}{\sqrt{1-t^2}}\ln t \\ &=\lim_{t\to1^-}t\cdot\lim_{t\to1^-}\frac{t-1}{\sqrt{1-t^2}}\\ &=1\cdot \lim_{t\to1^-}\sqrt{\frac{(t-1)^2}{1-t^2}}\\ &=\lim_{t\to1^-}\sqrt{\frac{1-t}{1+t}}\\ &=0 \end{aligned}\\ \therefore 由复合函数的极限运算法则知\\ \lim_{t\to1^-} e^{\tan x\ln t}=e^{\lim_{t\to1^-} {\tan x\ln t}}=e^0=1$$

法2

$$t=x-\frac\pi2\to0,x=\frac\pi2+t\\ \begin{aligned}\lim_{x\to\frac\pi2}(\sin x)^{\tan x} &=\lim_{x\to\frac\pi2}e^{\tan x\ln\sin x} \\&=\lim_{t\to0} \exp \left \{ \frac{\sin \left(\frac\pi2+t \right)\ln\left[\sin \left(\frac\pi2+t \right)\right] }{\cos \left(\frac\pi2+t \right)}\right\} \\&=\lim_{t\to0} \exp \left \{ \frac{\sin \left(\frac\pi2+t \right)\left[\sin \left(\frac\pi2+t \right)-1\right] }{\cos \left(\frac\pi2+t \right)}\right\} \\&=\lim_{t\to0} \exp \left \{ \frac{\cos t (\cos t-1) }{-\sin t}\right\} \\&=\lim_{t\to0} \exp \left \{ \frac{1\cdot(-\frac12t^2) }{-t}\right\} \\&=\lim_{t\to0} e^{\frac12t}=e^0=1 \end{aligned}$$

简化

$$\begin{aligned} \lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\tan x} &=\lim _{x \rightarrow 0}\left[\sin \left(\frac{\pi}{2}-x\right)\right]^{\tan \left(\frac{\pi}{2}-x\right)} \\ &=\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{\tan x}} \\ &=\exp \lim _{x \rightarrow 0} \frac{\cos x-1}{\tan x} \\ & = \exp \lim _{x \rightarrow 0} \frac{-x^{2} / 2}{x}=e^{0}=1 \end{aligned}$$ $$\lim_{x\to0}\frac{\cos(\tan x)-\cos x}{x^3\sin x}$$

法1

$$\begin{aligned} &\quad\,\lim_{x\to0}\frac{\cos(\tan x)-\cos x}{x^4}\\&= \lim_{x\to0}\frac{\cos(\tan x)-\cos x}{x^3\sin x}\\ &=\lim_{x\to0}\frac {\frac12(x^2-\tan^2x)+\frac1{24}(\tan^4x-x^4)+o(x^4)}{x^4}\\ &=\lim_{x\to0}\frac{(x-\tan x)2x[\frac12+\frac1{24} (\tan^2x+x^2)]+o(x^4)}{x^4} \\&=\frac{x(-\frac13x^3)}{x^4}=-\frac13 \end{aligned}$$

法2

$$f(x)=\cos x,\frac{\cos (\tan x)-\cos x}{\tan x-x}=f'(\xi),\xi\in(x,\tan x)\\ \begin{aligned} &\quad\,\lim_{x\to0}\frac{\cos(\tan x)-\cos x}{x^4}\\ &=\lim_{x\to0}\frac{(\tan x-x)f'(\xi)}{x^3\sin x}\\ &=\lim_{x\to0}\frac{(\tan x-x)(-\sin \xi)}{x^3\sin x}\\ &=\lim_{x\to0}-\frac{1}{3}\frac{\sin \xi}{\sin x}=-\frac 13\\ \end{aligned}$$ $$\lim_{n\to \infty}\frac1{n^4}\prod_{k=1}^{2n}(n^2+k^2)^{\frac1n}=\lim_{n\to \infty}A$$ $$\int_0^kf(x)\text{d}x=\lim_{n\to \infty}\frac{k}n\sum_{i=1}^nf\left(\frac {ki}n\right)$$ $$\begin{aligned}A&=\exp\left [\frac1n\sum_{k=1}^{2n}\ln(n^2+k^2)+2\ln\frac{1}{n^2}\right]\\&= \exp\left [\frac1n\sum_{k=1}^{2n}\ln(n^2+k^2)+2n\frac1{n}\ln\frac{1}{n^2}\right]\\ &=\exp\left [\frac1n\sum_{k=1}^{2n}\ln\left(\frac{n^2+k^2}{n^2}\right)\right]\\ &=\exp\left [\frac1n\sum_{k=1}^{2n}\ln\left(1+\frac{k^2}{n^2}\right)\right]\\ \lim_{n\to \infty}A&=\exp \int_0^2\ln(1+x^2)\text{d}x\\ &=\exp\left\{\left[x\ln(1+x^2)\right]_0^2-\int_0^2\frac{2x^2}{1+x^2}\text{d}x\right\} \\ &=e^{2(\ln5-2+\arctan 2)} \end{aligned}$$