高数第七版_习题解答_3-2 考研题提示及答案
3-2 考研題提示及答案
1、求極限
lim?x→0sin?x?sin?(sin?x)x3\lim_{x\rightarrow0}\frac{\sin x- \sin(\sin x)}{x^3} x→0lim?x3sinx?sin(sinx)?
=lim?x→0sin?x?sin?(sin?x)x3=L′Hospitallim?x→0(sin?x?sin?(sin?x))′(x3)′=lim?x→0cos?x?cos?(sin?x)?cos?x3x2=(lim?x→0cos?x)?(lim?x→01?cos?(sin?x)3x2)=L′Hospitallim?x→0x223x2=16\begin{aligned} &=\lim_{x\rightarrow0}\frac{\sin x- \sin(\sin x)}{x^3} \\ &\overset{{\color{red}L'Hospital}}{=} \lim_{x\rightarrow0}\frac{(\sin x- \sin(\sin x))'}{(x^3)'}\\ &= \lim_{x\rightarrow0}\frac{\cos x- \cos(\sin x)\cdot \cos x}{3 x^2}\\ &= \left(\lim_{x\rightarrow0} \cos x \right) \cdot\left( \lim_{x\rightarrow0}\frac{1- \cos(\sin x)}{3 x^2} \right)\\ &\overset{{\color{red}L'Hospital}}{=}\lim_{x\rightarrow0}\frac{\frac{x^2}{2}}{3 x^2}=\frac{1}{6} \end{aligned} ?=x→0lim?x3sinx?sin(sinx)?=L′Hospitalx→0lim?(x3)′(sinx?sin(sinx))′?=x→0lim?3x2cosx?cos(sinx)?cosx?=(x→0lim?cosx)?(x→0lim?3x21?cos(sinx)?)=L′Hospitalx→0lim?3x22x2??=61??
注：這里關鍵是靈活運用等價無窮小替換和洛必達法則
1?cos?(sin?x)～12(sin?x)2～12x2{\color{red}1-\cos (\sin x) \sim \frac{1}{2}(\sin x)^2\sim \frac{1}{2}x^2}1?cos(sinx)～21?(sinx)2～21?x2
2、求極限
lim?x→0(cos?x)1ln?(1+x2)\lim_{x\rightarrow0}(\cos x)^{\frac{1}{\ln (1+x^2)}} x→0lim?(cosx)ln(1+x2)1?
分析: 利用公式：
lim?(1+u)v=elim?(v?ln?(1+u))=elim?v?u\lim(1+u)^v = e^{ \lim (v\cdot\ln(1+u))}=e^{ \lim v\cdot u} lim(1+u)v=elim(v?ln(1+u))=elimv?u
于是：
lim?x→0(cos?x)1ln?(1+x2)=lim?x→0e1?2sin?2x2ln?(1+x2)=lim?x→0e?12=1e\lim_{x\rightarrow0}(\cos x)^{\frac{1}{\ln (1+x^2)}}=\lim_{x\rightarrow0}e^{\frac{1-2\sin^2\frac{x}{2}}{\ln(1+x^2)}}=\lim_{x\rightarrow0}e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}} x→0lim?(cosx)ln(1+x2)1?=x→0lim?eln(1+x2)1?2sin22x??=x→0lim?e?21?=e?1?
總結
以上是生活随笔為你收集整理的高数第七版_习题解答_3-2 考研题提示及答案的全部內容，希望文章能夠幫你解決所遇到的問題。
- 上一篇： 高数第七版_习题解答_3-1行列式习题
- 下一篇： 4行关键代码实现灰色模型GM(1, 1)