问个用定积分求极限限的题
点击联系发帖人
时间：2012-02-06 04:01
问一道求极限的题第2题,求极限结果等于多少呀?
因为两个重要极限limx->00(1+1/x)^x=e所以(（x+1)/x)^x=e所以(1/(x+1)^x=(1-x/(x+1)=1
为您推荐：
其他类似问题
是1，（1-1/（1+x））
扫描下载二维码大一高数求极限题！！求大神解答，求速度！大一高数求极限题！！求大神解答，求速度！如图问题都在题边写着-知识宝库
你可能对下面的信息感兴趣问个求极限的题_高等数学吧_百度贴吧
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&签到排名：今日本吧第个签到，本吧因你更精彩，明天继续来努力！
本吧签到人数：0成为超级会员，使用一键签到本月漏签0次！成为超级会员，赠送8张补签卡连续签到：天&&累计签到：天超级会员单次开通12个月以上，赠送连续签到卡3张
关注：188,282贴子：
问个求极限的题收藏
lim(x→π/3)sin(x-π/3)/(1-2cosx)
不要用求导的方法，我们还没学到，也不要用那啥法则。郁闷啊！老师几句话就说完了，同学们一个个的都说是，我没听明白啊！
高等数学 放心淘,超值淘,产品丰富,应有尽有,任你选购,畅享网购乐趣!淘宝网-淘,你喜欢!
三分之跟三？
呵呵…那就等你们老师做吧。这道题是要用洛必达法则做的，三分之跟三。
呵呵，会做了，不用罗比达法则也行。
登录百度帐号推荐应用
为兴趣而生，贴吧更懂你。或&img src=&/174aad22389bdf9c1ee61cae_b.png& data-rawwidth=&225& data-rawheight=&128& class=&content_image& width=&225&&xx考研老师说这个极限是等价不存在的，不等于1。&br&解释是这样说的：当x趋向于零时，sinx等价于x(等价无穷小)，然后x是趋向于0但x不能等于0，x*sin(1/x)是可能取到0的，因为sin(1/x)是振荡的，此时分母取到零点，就说明函数在这一点是无定义的，违背了函数f(x)在x-&0时处处有定义的说法，所以这个极限是不存在的。&br&但是用MATLAB计算是等于1的&br&&img src=&/d670c42a661c_b.png& data-rawwidth=&296& data-rawheight=&188& class=&content_image& width=&296&&&img src=&/9dd3fcfcf894e_b.jpg& data-rawwidth=&334& data-rawheight=&60& class=&content_image& width=&334&&&img src=&/9ebee32a39c4890c71ee_b.jpg& data-rawwidth=&561& data-rawheight=&496& class=&origin_image zh-lightbox-thumb& width=&561& data-original=&/9ebee32a39c4890c71ee_r.jpg&&&img src=&/fef5c728cdadac38c46a4aed9c19635f_b.jpg& data-rawwidth=&566& data-rawheight=&496& class=&origin_image zh-lightbox-thumb& width=&566& data-original=&/fef5c728cdadac38c46a4aed9c19635f_r.jpg&&所以感到很困惑，到底是怎么样的？
xx考研老师说这个极限是等价不存在的，不等于1。解释是这样说的：当x趋向于零时，sinx等价于x(等价无穷小)，然后x是趋向于0但x不能等于0，x*sin(1/x)是可能取到0的，因为sin(1/x)是振荡的，此时分母取到零点，就说明函数在这一点是无定义的，违背了函数f(x)在x-&0时处处有定义的说法，所以这个极限是不存在的。但是用MATLAB计算是等于1的…
闲人一枚，试着用Mathematica做了一下：下面把图像尽量放大：下面把图像尽量放大：（好没有帮助的回答→_→）
谢邀。以前回答过这个问题：&a href=&/question//answer/& class=&internal&&当x趋向于0，sin(xsin1/x)/(xsin1/x)是否为1？ - 余翔的回答&/a&，不过那个问题好像被举报了。下面是以前的回答，修订了一下错误。&br&&br&是的，不过更准确的应该是&img src=&///equation?tex=%5C%28%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+E%5Csetminus%5C%7B0%5C%7D%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+%3D1%5C%29& alt=&\(\lim_{x \rightarrow 0;x\in E\setminus\{0\}} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} =1\)& eeimg=&1&&。&br&首先需要知道极限&img src=&///equation?tex=%5C%28%5Clim_%7Bx+%5Crightarrow+0%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+%5C%29& alt=&\(\lim_{x \rightarrow 0} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} \)& eeimg=&1&&指的是什么？因为&img src=&///equation?tex=x& alt=&x& eeimg=&1&&并不能取所有的实数，有些点必须排除。要使得函数&img src=&///equation?tex=%5Ctextstyle+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+& alt=&\textstyle \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} & eeimg=&1&&有意义，分母&img src=&///equation?tex=%5Ctextstyle+x%5Csin%5Cfrac%7B1%7D%7Bx%7D+%5Cneq+0& alt=&\textstyle x\sin\frac{1}{x} \neq 0& eeimg=&1&&，于是&img src=&///equation?tex=%5Ctextstyle+x%5Cneq0%2C%5Cpm%5Cfrac%7B1%7D%7B%5Cpi%7D%2C%5Cpm%5Cfrac%7B1%7D%7B2%5Cpi%7D%2C%5Cdots& alt=&\textstyle x\neq0,\pm\frac{1}{\pi},\pm\frac{1}{2\pi},\dots& eeimg=&1&&&br&设集合&img src=&///equation?tex=%5Ctextstyle+E%3A%3D%5C%7Bx%5Cin%5Cmathbf%7BR%7D%3Ax%5Cneq0%2C%5Cpm%5Cfrac%7B1%7D%7B%5Cpi%7D%2C%5Cpm%5Cfrac%7B1%7D%7B2%5Cpi%7D+%2C%5Ccdots%5C%7D& alt=&\textstyle E:=\{x\in\mathbf{R}:x\neq0,\pm\frac{1}{\pi},\pm\frac{1}{2\pi} ,\cdots\}& eeimg=&1&&，那么函数&img src=&///equation?tex=%5Ctextstyle+x%5Cmapsto+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+& alt=&\textstyle x\mapsto \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} & eeimg=&1&&的&b&定义域&/b&为&img src=&///equation?tex=E& alt=&E& eeimg=&1&&。极限&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+& alt=&\textstyle \lim_{x \rightarrow 0} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} & eeimg=&1&&指的是&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+E%5Cbackslash%5C%7B0%5C%7D%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+& alt=&\textstyle \lim_{x \rightarrow 0;x\in E\backslash\{0\}} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} & eeimg=&1&&（注意&img src=&///equation?tex=0& alt=&0& eeimg=&1&&是&img src=&///equation?tex=E& alt=&E& eeimg=&1&&的极限点，因此这个极限是定义良好的，由于&img src=&///equation?tex=E& alt=&E& eeimg=&1&&不包含&img src=&///equation?tex=0& alt=&0& eeimg=&1&&，所以&img src=&///equation?tex=E%5Csetminus%5C%7B0%5C%7D& alt=&E\setminus\{0\}& eeimg=&1&&和&img src=&///equation?tex=E& alt=&E& eeimg=&1&&没有区别，但更一般的情形会有区别）。&br&&br&先回顾一下函数极限的定义&br&&b&定义.&/b&（函数在一点处的极限） 设&img src=&///equation?tex=X& alt=&X& eeimg=&1&&是&img src=&///equation?tex=%5Cmathbf%7BR%7D& alt=&\mathbf{R}& eeimg=&1&&子集，&img src=&///equation?tex=f%3AX%5Cto%5Cmathbf%7BR%7D& alt=&f:X\to\mathbf{R}& eeimg=&1&&是函数，并设&img src=&///equation?tex=E& alt=&E& eeimg=&1&&是&img src=&///equation?tex=X& alt=&X& eeimg=&1&&的子集，&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&是&img src=&///equation?tex=E& alt=&E& eeimg=&1&&的&a href=&///?target=https%3A//en.wikipedia.org/wiki/Limit_point& class=& wrap external& target=&_blank& rel=&nofollow noreferrer&&极限点&i class=&icon-external&&&/i&&/a&，而&img src=&///equation?tex=L& alt=&L& eeimg=&1&&是实数，我们说&img src=&///equation?tex=f& alt=&f& eeimg=&1&&在&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&处沿着&b&&img src=&///equation?tex=E& alt=&E& eeimg=&1&&收敛&/b&到&img src=&///equation?tex=L& alt=&L& eeimg=&1&&，写作&br&&img src=&///equation?tex=%5Clim_%7Bx+%5Crightarrow+x_0%3Bx%5Cin+E%5Csetminus%5C%7Bx_0%5C%7D%7D%7Bf%28x%29%7D+%3DL& alt=&\lim_{x \rightarrow x_0;x\in E\setminus\{x_0\}}{f(x)} =L& eeimg=&1&&&br&当且仅当对于每个&img src=&///equation?tex=%5Cvarepsilon%3E0& alt=&\varepsilon&0& eeimg=&1&&，都存在&img src=&///equation?tex=%5Cdelta%3E0& alt=&\delta&0& eeimg=&1&&，对于一切&img src=&///equation?tex=x%5Cin+E%5Cbackslash%5C%7Bx_0%5C%7D& alt=&x\in E\backslash\{x_0\}& eeimg=&1&&，当&img src=&///equation?tex=%7Cx-x_0%7C%5Cleq+%5Cdelta& alt=&|x-x_0|\leq \delta& eeimg=&1&&时，&img src=&///equation?tex=%7Cf%28x%29-L%7C%5Cleq+%5Cvarepsilon& alt=&|f(x)-L|\leq \varepsilon& eeimg=&1&&。&br&&br&&b&注. &/b&我们只考虑&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&是&img src=&///equation?tex=E& alt=&E& eeimg=&1&&的极限点时，函数在&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&处的极限，当&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&不是极限点时，不值得定义极限的概念（为什么？）。很多情况我们从上面的记号中略去集合&img src=&///equation?tex=E%5Csetminus%5C%7Bx_0%5C%7D& alt=&E\setminus\{x_0\}& eeimg=&1&&，也就是说，我们只说&img src=&///equation?tex=f& alt=&f& eeimg=&1&&在&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&处收敛到&img src=&///equation?tex=L& alt=&L& eeimg=&1&&，或者&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+x_0%7D%7Bf%28x%29%7D+%3DL& alt=&\textstyle \lim_{x \rightarrow x_0}{f(x)} =L& eeimg=&1&&，但去掉集合&img src=&///equation?tex=E%5Csetminus%5C%7Bx_0%5C%7D& alt=&E\setminus\{x_0\}& eeimg=&1&&有点危险，比如对于&a href=&///?target=http%3A///DirichletFunction.html& class=& wrap external& target=&_blank& rel=&nofollow noreferrer&&Dirichlet Function &i class=&icon-external&&&/i&&/a&&br&&img src=&///equation?tex=D%28x%29%3A%3D%5Cbegin%7Bcases%7D%0A1%2C+%26%5C+x%5Cin%5Cmathbf%7BQ%7D%5Ccap%5B0%2C1%5D%3B%5C%5C%0A0%2C%26%5C+x%5Cin+%5B0%2C1%5D+%5Csetminus%5Cmathbf%7BQ%7D+.%0A%5Cend%7Bcases%7D& alt=&D(x):=\begin{cases}
1, &\ x\in\mathbf{Q}\cap[0,1];\\
0,&\ x\in [0,1] \setminus\mathbf{Q} .
\end{cases}& eeimg=&1&&&br&极限&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+%5B0%2C1%5D%5Csetminus%5C%7B0%5C%7D%7D+D%28x%29& alt=&\textstyle \lim_{x \rightarrow 0;x\in [0,1]\setminus\{0\}} D(x)& eeimg=&1&&不存在，但极限&br&&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+%5Cmathbf%7BQ%7D%5Ccap%5B0%2C1%5D%5Csetminus%5C%7B0%5C%7D%7D+D%28x%29%3D1& alt=&\textstyle \lim_{x \rightarrow 0;x\in \mathbf{Q}\cap[0,1]\setminus\{0\}} D(x)=1& eeimg=&1&&，&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+%28%5B0%2C1%5D%5Csetminus+%5Cmathbf%7BQ%7D%29%5Csetminus%5C%7B0%5C%7D%7D+D%28x%29%3D0& alt=&\textstyle \lim_{x \rightarrow 0;x\in ([0,1]\setminus \mathbf{Q})\setminus\{0\}} D(x)=0& eeimg=&1&&&br&都存在&br&&br&函数极限可以用序列极限刻画，因为我们有下面命题&br&&b&命题.&/b& 设&img src=&///equation?tex=X& alt=&X& eeimg=&1&&是&img src=&///equation?tex=%5Cmathbf%7BR%7D& alt=&\mathbf{R}& eeimg=&1&&的子集，&img src=&///equation?tex=f%3AX%5Cto%5Cmathbf%7BR%7D& alt=&f:X\to\mathbf{R}& eeimg=&1&&是函数，并设&img src=&///equation?tex=E& alt=&E& eeimg=&1&&是&img src=&///equation?tex=X& alt=&X& eeimg=&1&&的子集，&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&是&img src=&///equation?tex=E& alt=&E& eeimg=&1&&的&b&极限点&/b&，而&img src=&///equation?tex=L& alt=&L& eeimg=&1&&是实数，那么下述两个命题是逻辑上等价的：&br&&ul&&li&&img src=&///equation?tex=f& alt=&f& eeimg=&1&&在&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&处沿着&img src=&///equation?tex=E& alt=&E& eeimg=&1&&收敛到&img src=&///equation?tex=L& alt=&L& eeimg=&1&&&br&&/li&&li&对于每个完全由&img src=&///equation?tex=E%5Cbackslash%5C%7Bx_0%5C%7D& alt=&E\backslash\{x_0\}& eeimg=&1&&的元素组成，并且收敛到&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&的序列&img src=&///equation?tex=%28a_n%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(a_n)_{n=0}^{\infty}& eeimg=&1&&，序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=L& alt=&L& eeimg=&1&&&br&&/li&&/ul&&br&现在采用序列来证明极限&img src=&///equation?tex=%5C%28%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+E%5Csetminus%5C%7B0%5C%7D%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+%3D1%5C%29& alt=&\(\lim_{x \rightarrow 0;x\in E\setminus\{0\}} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} =1\)& eeimg=&1&&。设&img src=&///equation?tex=%5Ctextstyle+f%28x%29%3Dx%5Csin%5Cfrac%7B1%7D%7Bx%7D+& alt=&\textstyle f(x)=x\sin\frac{1}{x} & eeimg=&1&&，&img src=&///equation?tex=%5Ctextstyle+g%28x%29%3D%5Cfrac%7B%5Csin+x%7D%7Bx%7D& alt=&\textstyle g(x)=\frac{\sin x}{x}& eeimg=&1&&，根据这一命题，由于&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+E%5Cbackslash%5C%7B0%5C%7D%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D+%7D+%3D0& alt=&\textstyle \lim_{x \rightarrow 0;x\in E\backslash\{0\}}{x\sin\frac{1}{x} } =0& eeimg=&1&&，于是每个完全由&img src=&///equation?tex=E%5Cbackslash%5C%7B0%5C%7D& alt=&E\backslash\{0\}& eeimg=&1&&的元素组成，并且收敛到&img src=&///equation?tex=0& alt=&0& eeimg=&1&&的序列&img src=&///equation?tex=%28a_n%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(a_n)_{n=0}^{\infty}& eeimg=&1&&，即&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bn+%5Crightarrow+%5Cinfty%7D%7Ba_n%7D+%3D0& alt=&\textstyle \lim_{n \rightarrow \infty}{a_n} =0& eeimg=&1&&，序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=0& alt=&0& eeimg=&1&&，且对于每个&img src=&///equation?tex=n& alt=&n& eeimg=&1&&，&img src=&///equation?tex=f%28a_n%29%5Cneq+0& alt=&f(a_n)\neq 0& eeimg=&1&&，因为&img src=&///equation?tex=a_n%5Cin+E%5Cbackslash%5C%7Bx_0%5C%7D& alt=&a_n\in E\backslash\{x_0\}& eeimg=&1&&。又因为&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+%5Cmathbf%7BR%7D%5Csetminus%5C%7B0%5C%7D%7D%7B%5Cfrac%7B%5Csin+x%7D%7Bx%7D+%7D+%3D1& alt=&\textstyle \lim_{x \rightarrow 0;x\in \mathbf{R}\setminus\{0\}}{\frac{\sin x}{x} } =1& eeimg=&1&&，而序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=0& alt=&0& eeimg=&1&&且每一项都不为&img src=&///equation?tex=0& alt=&0& eeimg=&1&&，即对每个&img src=&///equation?tex=n& alt=&n& eeimg=&1&&，&img src=&///equation?tex=f%28a_n%29%5Cin+%5Cmathbf%7BR%7D%5Cbackslash%5C%7B0%5C%7D& alt=&f(a_n)\in \mathbf{R}\backslash\{0\}& eeimg=&1&&，于是序列&img src=&///equation?tex=%28g%28f%28a_n%29+%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(g(f(a_n) )_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=1& alt=&1& eeimg=&1&&，也就是序列&img src=&///equation?tex=%5Ctextstyle+%5Cleft%28%5Cfrac%7B%5Csin%28a_n%5Csin%5Cfrac%7B1%7D%7Ba_n%7D%29+%7D%7Ba_n%5Csin%5Cfrac%7B1%7D%7Ba_n%7D%7D+%5Cright%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&\textstyle \left(\frac{\sin(a_n\sin\frac{1}{a_n}) }{a_n\sin\frac{1}{a_n}} \right)_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=1%0A& alt=&1
& eeimg=&1&&，因此函数&img src=&///equation?tex=%5Ctextstyle+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+& alt=&\textstyle \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} & eeimg=&1&&在&img src=&///equation?tex=0& alt=&0& eeimg=&1&&处沿着&img src=&///equation?tex=E& alt=&E& eeimg=&1&&收敛到&img src=&///equation?tex=1& alt=&1& eeimg=&1&&，即&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+E%5Cbackslash%5C%7B0%5C%7D%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+%3D1& alt=&\textstyle \lim_{x \rightarrow 0;x\in E\backslash\{0\}} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} =1& eeimg=&1&&&br&&br&注意，下面命题&b&不成立&/b&&br&&b&命题. &/b&（&b&不成立）&/b&设&img src=&///equation?tex=X%2CY& alt=&X,Y& eeimg=&1&&是&img src=&///equation?tex=%5Cmathbf%7BR%7D& alt=&\mathbf{R}& eeimg=&1&&的子集，设&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&是&img src=&///equation?tex=X& alt=&X& eeimg=&1&&的极限点，&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&是&img src=&///equation?tex=Y& alt=&Y& eeimg=&1&&的极限点，设&img src=&///equation?tex=f%3AX%5Cto+Y& alt=&f:X\to Y& eeimg=&1&&是函数，使得&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+x_0%3Bx%5Cin+X%5Cbackslash%5C%7Bx_0%5C%7D%7D%7Bf%28x%29%7D+%3Dy_0& alt=&\textstyle \lim_{x \rightarrow x_0;x\in X\backslash\{x_0\}}{f(x)} =y_0& eeimg=&1&&&br&，设&img src=&///equation?tex=g%3AY%5Cto+%5Cmathbf%7BR%7D& alt=&g:Y\to \mathbf{R}& eeimg=&1&&是函数，使得&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7By+%5Crightarrow+y_0%3Bx%5Cin+Y%5Cbackslash%5C%7By_0%5C%7D%7D%7Bg%28y%29%7D+%3Dz_0& alt=&\textstyle \lim_{y \rightarrow y_0;x\in Y\backslash\{y_0\}}{g(y)} =z_0& eeimg=&1&&，那么&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+x_0%3Bx%5Cin+X%5Cbackslash%5C%7Bx_0%5C%7D%7D%7Bg%28f%28x%29%29%7D+%3Dz_0& alt=&\textstyle \lim_{x \rightarrow x_0;x\in X\backslash\{x_0\}}{g(f(x))} =z_0& eeimg=&1&&&br&这个命题是&b&错误&/b&的，因此不能直接根据&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D+%7D+%3D0& alt=&\textstyle \lim_{x \rightarrow 0}{x\sin\frac{1}{x} } =0& eeimg=&1&&和&img src=&///equation?tex=+%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%7D%7B%5Cfrac%7B%5Csin+x%7D%7Bx%7D+%7D+%3D1& alt=& \textstyle \lim_{x \rightarrow 0}{\frac{\sin x}{x} } =1& eeimg=&1&&来说明&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+%3D1& alt=&\textstyle \lim_{x \rightarrow 0} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} =1& eeimg=&1&&&br&&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+x_0%3Bx%5Cin+X%5Cbackslash%5C%7Bx_0%5C%7D%7D%7Bg%28f%28x%29%29%7D& alt=&\textstyle \lim_{x \rightarrow x_0;x\in X\backslash\{x_0\}}{g(f(x))}& eeimg=&1&&不一定等于&img src=&///equation?tex=z_0& alt=&z_0& eeimg=&1&&。但如果函数&img src=&///equation?tex=g& alt=&g& eeimg=&1&&在&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&处连续，这个命题是成立的。因此可以通过补充定义&img src=&///equation?tex=%5Ctfrac%7B%5Csin%28x%29%7D%7Bx%7D& alt=&\tfrac{\sin(x)}{x}& eeimg=&1&&在&img src=&///equation?tex=0& alt=&0& eeimg=&1&&处的值使得它连续，这样就可以使用上述命题说明&img src=&///equation?tex=%5C%28%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+E%5Csetminus%5C%7B0%5C%7D%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+%3D1%5C%29& alt=&\(\lim_{x \rightarrow 0;x\in E\setminus\{0\}} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} =1\)& eeimg=&1&&。&br&&br&因为对于每个完全由&img src=&///equation?tex=X%5Cbackslash%5C%7B0%5C%7D& alt=&X\backslash\{0\}& eeimg=&1&&的元素组成，并且收敛到&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&的序列&img src=&///equation?tex=%28a_n%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(a_n)_{n=0}^{\infty}& eeimg=&1&&，序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&，但序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&可能有&b&无限&/b&多项等于&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&，而&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7By+%5Crightarrow+y_0%3By%5Cin+Y%5Cbackslash%5C%7By_0%5C%7D%7D%7Bg%28y%29%7D+%3Dz_0& alt=&\textstyle \lim_{y \rightarrow y_0;y\in Y\backslash\{y_0\}}{g(y)} =z_0& eeimg=&1&&只能说明对于每个完全由&img src=&///equation?tex=Y%5Cbackslash%5C%7B0%5C%7D& alt=&Y\backslash\{0\}& eeimg=&1&&的元素组成，并且收敛到&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&的序列&img src=&///equation?tex=%28b_n%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(b_n)_{n=0}^{\infty}& eeimg=&1&&，序列&img src=&///equation?tex=%28g%28b_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(g(b_n))_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=z_0& alt=&z_0& eeimg=&1&&，当序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&有无限多项等于&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&，我们不能推出序列&img src=&///equation?tex=%28g%28f%28a_n%29%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(g(f(a_n)))_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=z_0& alt=&z_0& eeimg=&1&&，但下面两种情况可以&br&&ul&&li&对于每个收敛到&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&的序列&img src=&///equation?tex=%28a_n%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(a_n)_{n=0}^{\infty}& eeimg=&1&&，序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&只有有限多项等于&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&，那么就有&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+x_0%3Bx%5Cin+X%5Cbackslash%5C%7Bx_0%5C%7D%7D%7Bg%28f%28x%29%29%7D+%3Dz_0& alt=&\textstyle \lim_{x \rightarrow x_0;x\in X\backslash\{x_0\}}{g(f(x))} =z_0& eeimg=&1&&&br&&/li&&li&函数&img src=&///equation?tex=g%28y%29& alt=&g(y)& eeimg=&1&&在&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&处&b&连续&/b&时，我们也有&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+x_0%3Bx%5Cin+X%5Cbackslash%5C%7Bx_0%5C%7D%7D%7Bg%28f%28x%29%29%7D+%3Dz_0& alt=&\textstyle \lim_{x \rightarrow x_0;x\in X\backslash\{x_0\}}{g(f(x))} =z_0& eeimg=&1&&。&/li&&/ul&
谢邀。以前回答过这个问题：，不过那个问题好像被举报了。下面是以前的回答，修订了一下错误。 是的，不过更准确的应该是\(\lim_{x \rightarrow 0;x\in E\setminus\{0\}} \frac{\sin(x\sin\frac{1…
我觉得你老师的意思是，x_k=1/(kpi)的时候分母是0，式子没定义，而x_k是一个收敛到0的序列，所以极限不是well-defined的。&br&个人觉得这个解释make sense，但是问题本身属于没什么价值的细枝末节，没太多纠结的必要。
我觉得你老师的意思是，x_k=1/(kpi)的时候分母是0，式子没定义，而x_k是一个收敛到0的序列，所以极限不是well-defined的。 个人觉得这个解释make sense，但是问题本身属于没什么价值的细枝末节，没太多纠结的必要。
已有帐号？
无法登录？
社交帐号登录
白痴一只。}
因为两个重要极限limx->00(1+1/x)^x=e所以(（x+1)/x)^x=e所以(1/(x+1)^x=(1-x/(x+1)=1
为您推荐：
其他类似问题
是1，（1-1/（1+x））
扫描下载二维码大一高数求极限题！！求大神解答，求速度！大一高数求极限题！！求大神解答，求速度！如图问题都在题边写着-知识宝库
你可能对下面的信息感兴趣问个求极限的题_高等数学吧_百度贴吧
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&签到排名：今日本吧第个签到，本吧因你更精彩，明天继续来努力！
本吧签到人数：0成为超级会员，使用一键签到本月漏签0次！成为超级会员，赠送8张补签卡连续签到：天&&累计签到：天超级会员单次开通12个月以上，赠送连续签到卡3张
关注：188,282贴子：
问个求极限的题收藏
lim(x→π/3)sin(x-π/3)/(1-2cosx)
不要用求导的方法，我们还没学到，也不要用那啥法则。郁闷啊！老师几句话就说完了，同学们一个个的都说是，我没听明白啊！
高等数学 放心淘,超值淘,产品丰富,应有尽有,任你选购,畅享网购乐趣!淘宝网-淘,你喜欢!
三分之跟三？
呵呵…那就等你们老师做吧。这道题是要用洛必达法则做的，三分之跟三。
呵呵，会做了，不用罗比达法则也行。
登录百度帐号推荐应用
为兴趣而生，贴吧更懂你。或&img src=&/174aad22389bdf9c1ee61cae_b.png& data-rawwidth=&225& data-rawheight=&128& class=&content_image& width=&225&&xx考研老师说这个极限是等价不存在的，不等于1。&br&解释是这样说的：当x趋向于零时，sinx等价于x(等价无穷小)，然后x是趋向于0但x不能等于0，x*sin(1/x)是可能取到0的，因为sin(1/x)是振荡的，此时分母取到零点，就说明函数在这一点是无定义的，违背了函数f(x)在x-&0时处处有定义的说法，所以这个极限是不存在的。&br&但是用MATLAB计算是等于1的&br&&img src=&/d670c42a661c_b.png& data-rawwidth=&296& data-rawheight=&188& class=&content_image& width=&296&&&img src=&/9dd3fcfcf894e_b.jpg& data-rawwidth=&334& data-rawheight=&60& class=&content_image& width=&334&&&img src=&/9ebee32a39c4890c71ee_b.jpg& data-rawwidth=&561& data-rawheight=&496& class=&origin_image zh-lightbox-thumb& width=&561& data-original=&/9ebee32a39c4890c71ee_r.jpg&&&img src=&/fef5c728cdadac38c46a4aed9c19635f_b.jpg& data-rawwidth=&566& data-rawheight=&496& class=&origin_image zh-lightbox-thumb& width=&566& data-original=&/fef5c728cdadac38c46a4aed9c19635f_r.jpg&&所以感到很困惑，到底是怎么样的？
xx考研老师说这个极限是等价不存在的，不等于1。解释是这样说的：当x趋向于零时，sinx等价于x(等价无穷小)，然后x是趋向于0但x不能等于0，x*sin(1/x)是可能取到0的，因为sin(1/x)是振荡的，此时分母取到零点，就说明函数在这一点是无定义的，违背了函数f(x)在x-&0时处处有定义的说法，所以这个极限是不存在的。但是用MATLAB计算是等于1的…
闲人一枚，试着用Mathematica做了一下：下面把图像尽量放大：下面把图像尽量放大：（好没有帮助的回答→_→）
谢邀。以前回答过这个问题：&a href=&/question//answer/& class=&internal&&当x趋向于0，sin(xsin1/x)/(xsin1/x)是否为1？ - 余翔的回答&/a&，不过那个问题好像被举报了。下面是以前的回答，修订了一下错误。&br&&br&是的，不过更准确的应该是&img src=&///equation?tex=%5C%28%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+E%5Csetminus%5C%7B0%5C%7D%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+%3D1%5C%29& alt=&\(\lim_{x \rightarrow 0;x\in E\setminus\{0\}} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} =1\)& eeimg=&1&&。&br&首先需要知道极限&img src=&///equation?tex=%5C%28%5Clim_%7Bx+%5Crightarrow+0%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+%5C%29& alt=&\(\lim_{x \rightarrow 0} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} \)& eeimg=&1&&指的是什么？因为&img src=&///equation?tex=x& alt=&x& eeimg=&1&&并不能取所有的实数，有些点必须排除。要使得函数&img src=&///equation?tex=%5Ctextstyle+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+& alt=&\textstyle \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} & eeimg=&1&&有意义，分母&img src=&///equation?tex=%5Ctextstyle+x%5Csin%5Cfrac%7B1%7D%7Bx%7D+%5Cneq+0& alt=&\textstyle x\sin\frac{1}{x} \neq 0& eeimg=&1&&，于是&img src=&///equation?tex=%5Ctextstyle+x%5Cneq0%2C%5Cpm%5Cfrac%7B1%7D%7B%5Cpi%7D%2C%5Cpm%5Cfrac%7B1%7D%7B2%5Cpi%7D%2C%5Cdots& alt=&\textstyle x\neq0,\pm\frac{1}{\pi},\pm\frac{1}{2\pi},\dots& eeimg=&1&&&br&设集合&img src=&///equation?tex=%5Ctextstyle+E%3A%3D%5C%7Bx%5Cin%5Cmathbf%7BR%7D%3Ax%5Cneq0%2C%5Cpm%5Cfrac%7B1%7D%7B%5Cpi%7D%2C%5Cpm%5Cfrac%7B1%7D%7B2%5Cpi%7D+%2C%5Ccdots%5C%7D& alt=&\textstyle E:=\{x\in\mathbf{R}:x\neq0,\pm\frac{1}{\pi},\pm\frac{1}{2\pi} ,\cdots\}& eeimg=&1&&，那么函数&img src=&///equation?tex=%5Ctextstyle+x%5Cmapsto+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+& alt=&\textstyle x\mapsto \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} & eeimg=&1&&的&b&定义域&/b&为&img src=&///equation?tex=E& alt=&E& eeimg=&1&&。极限&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+& alt=&\textstyle \lim_{x \rightarrow 0} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} & eeimg=&1&&指的是&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+E%5Cbackslash%5C%7B0%5C%7D%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+& alt=&\textstyle \lim_{x \rightarrow 0;x\in E\backslash\{0\}} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} & eeimg=&1&&（注意&img src=&///equation?tex=0& alt=&0& eeimg=&1&&是&img src=&///equation?tex=E& alt=&E& eeimg=&1&&的极限点，因此这个极限是定义良好的，由于&img src=&///equation?tex=E& alt=&E& eeimg=&1&&不包含&img src=&///equation?tex=0& alt=&0& eeimg=&1&&，所以&img src=&///equation?tex=E%5Csetminus%5C%7B0%5C%7D& alt=&E\setminus\{0\}& eeimg=&1&&和&img src=&///equation?tex=E& alt=&E& eeimg=&1&&没有区别，但更一般的情形会有区别）。&br&&br&先回顾一下函数极限的定义&br&&b&定义.&/b&（函数在一点处的极限） 设&img src=&///equation?tex=X& alt=&X& eeimg=&1&&是&img src=&///equation?tex=%5Cmathbf%7BR%7D& alt=&\mathbf{R}& eeimg=&1&&子集，&img src=&///equation?tex=f%3AX%5Cto%5Cmathbf%7BR%7D& alt=&f:X\to\mathbf{R}& eeimg=&1&&是函数，并设&img src=&///equation?tex=E& alt=&E& eeimg=&1&&是&img src=&///equation?tex=X& alt=&X& eeimg=&1&&的子集，&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&是&img src=&///equation?tex=E& alt=&E& eeimg=&1&&的&a href=&///?target=https%3A//en.wikipedia.org/wiki/Limit_point& class=& wrap external& target=&_blank& rel=&nofollow noreferrer&&极限点&i class=&icon-external&&&/i&&/a&，而&img src=&///equation?tex=L& alt=&L& eeimg=&1&&是实数，我们说&img src=&///equation?tex=f& alt=&f& eeimg=&1&&在&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&处沿着&b&&img src=&///equation?tex=E& alt=&E& eeimg=&1&&收敛&/b&到&img src=&///equation?tex=L& alt=&L& eeimg=&1&&，写作&br&&img src=&///equation?tex=%5Clim_%7Bx+%5Crightarrow+x_0%3Bx%5Cin+E%5Csetminus%5C%7Bx_0%5C%7D%7D%7Bf%28x%29%7D+%3DL& alt=&\lim_{x \rightarrow x_0;x\in E\setminus\{x_0\}}{f(x)} =L& eeimg=&1&&&br&当且仅当对于每个&img src=&///equation?tex=%5Cvarepsilon%3E0& alt=&\varepsilon&0& eeimg=&1&&，都存在&img src=&///equation?tex=%5Cdelta%3E0& alt=&\delta&0& eeimg=&1&&，对于一切&img src=&///equation?tex=x%5Cin+E%5Cbackslash%5C%7Bx_0%5C%7D& alt=&x\in E\backslash\{x_0\}& eeimg=&1&&，当&img src=&///equation?tex=%7Cx-x_0%7C%5Cleq+%5Cdelta& alt=&|x-x_0|\leq \delta& eeimg=&1&&时，&img src=&///equation?tex=%7Cf%28x%29-L%7C%5Cleq+%5Cvarepsilon& alt=&|f(x)-L|\leq \varepsilon& eeimg=&1&&。&br&&br&&b&注. &/b&我们只考虑&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&是&img src=&///equation?tex=E& alt=&E& eeimg=&1&&的极限点时，函数在&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&处的极限，当&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&不是极限点时，不值得定义极限的概念（为什么？）。很多情况我们从上面的记号中略去集合&img src=&///equation?tex=E%5Csetminus%5C%7Bx_0%5C%7D& alt=&E\setminus\{x_0\}& eeimg=&1&&，也就是说，我们只说&img src=&///equation?tex=f& alt=&f& eeimg=&1&&在&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&处收敛到&img src=&///equation?tex=L& alt=&L& eeimg=&1&&，或者&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+x_0%7D%7Bf%28x%29%7D+%3DL& alt=&\textstyle \lim_{x \rightarrow x_0}{f(x)} =L& eeimg=&1&&，但去掉集合&img src=&///equation?tex=E%5Csetminus%5C%7Bx_0%5C%7D& alt=&E\setminus\{x_0\}& eeimg=&1&&有点危险，比如对于&a href=&///?target=http%3A///DirichletFunction.html& class=& wrap external& target=&_blank& rel=&nofollow noreferrer&&Dirichlet Function &i class=&icon-external&&&/i&&/a&&br&&img src=&///equation?tex=D%28x%29%3A%3D%5Cbegin%7Bcases%7D%0A1%2C+%26%5C+x%5Cin%5Cmathbf%7BQ%7D%5Ccap%5B0%2C1%5D%3B%5C%5C%0A0%2C%26%5C+x%5Cin+%5B0%2C1%5D+%5Csetminus%5Cmathbf%7BQ%7D+.%0A%5Cend%7Bcases%7D& alt=&D(x):=\begin{cases}
1, &\ x\in\mathbf{Q}\cap[0,1];\\
0,&\ x\in [0,1] \setminus\mathbf{Q} .
\end{cases}& eeimg=&1&&&br&极限&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+%5B0%2C1%5D%5Csetminus%5C%7B0%5C%7D%7D+D%28x%29& alt=&\textstyle \lim_{x \rightarrow 0;x\in [0,1]\setminus\{0\}} D(x)& eeimg=&1&&不存在，但极限&br&&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+%5Cmathbf%7BQ%7D%5Ccap%5B0%2C1%5D%5Csetminus%5C%7B0%5C%7D%7D+D%28x%29%3D1& alt=&\textstyle \lim_{x \rightarrow 0;x\in \mathbf{Q}\cap[0,1]\setminus\{0\}} D(x)=1& eeimg=&1&&，&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+%28%5B0%2C1%5D%5Csetminus+%5Cmathbf%7BQ%7D%29%5Csetminus%5C%7B0%5C%7D%7D+D%28x%29%3D0& alt=&\textstyle \lim_{x \rightarrow 0;x\in ([0,1]\setminus \mathbf{Q})\setminus\{0\}} D(x)=0& eeimg=&1&&&br&都存在&br&&br&函数极限可以用序列极限刻画，因为我们有下面命题&br&&b&命题.&/b& 设&img src=&///equation?tex=X& alt=&X& eeimg=&1&&是&img src=&///equation?tex=%5Cmathbf%7BR%7D& alt=&\mathbf{R}& eeimg=&1&&的子集，&img src=&///equation?tex=f%3AX%5Cto%5Cmathbf%7BR%7D& alt=&f:X\to\mathbf{R}& eeimg=&1&&是函数，并设&img src=&///equation?tex=E& alt=&E& eeimg=&1&&是&img src=&///equation?tex=X& alt=&X& eeimg=&1&&的子集，&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&是&img src=&///equation?tex=E& alt=&E& eeimg=&1&&的&b&极限点&/b&，而&img src=&///equation?tex=L& alt=&L& eeimg=&1&&是实数，那么下述两个命题是逻辑上等价的：&br&&ul&&li&&img src=&///equation?tex=f& alt=&f& eeimg=&1&&在&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&处沿着&img src=&///equation?tex=E& alt=&E& eeimg=&1&&收敛到&img src=&///equation?tex=L& alt=&L& eeimg=&1&&&br&&/li&&li&对于每个完全由&img src=&///equation?tex=E%5Cbackslash%5C%7Bx_0%5C%7D& alt=&E\backslash\{x_0\}& eeimg=&1&&的元素组成，并且收敛到&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&的序列&img src=&///equation?tex=%28a_n%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(a_n)_{n=0}^{\infty}& eeimg=&1&&，序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=L& alt=&L& eeimg=&1&&&br&&/li&&/ul&&br&现在采用序列来证明极限&img src=&///equation?tex=%5C%28%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+E%5Csetminus%5C%7B0%5C%7D%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+%3D1%5C%29& alt=&\(\lim_{x \rightarrow 0;x\in E\setminus\{0\}} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} =1\)& eeimg=&1&&。设&img src=&///equation?tex=%5Ctextstyle+f%28x%29%3Dx%5Csin%5Cfrac%7B1%7D%7Bx%7D+& alt=&\textstyle f(x)=x\sin\frac{1}{x} & eeimg=&1&&，&img src=&///equation?tex=%5Ctextstyle+g%28x%29%3D%5Cfrac%7B%5Csin+x%7D%7Bx%7D& alt=&\textstyle g(x)=\frac{\sin x}{x}& eeimg=&1&&，根据这一命题，由于&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+E%5Cbackslash%5C%7B0%5C%7D%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D+%7D+%3D0& alt=&\textstyle \lim_{x \rightarrow 0;x\in E\backslash\{0\}}{x\sin\frac{1}{x} } =0& eeimg=&1&&，于是每个完全由&img src=&///equation?tex=E%5Cbackslash%5C%7B0%5C%7D& alt=&E\backslash\{0\}& eeimg=&1&&的元素组成，并且收敛到&img src=&///equation?tex=0& alt=&0& eeimg=&1&&的序列&img src=&///equation?tex=%28a_n%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(a_n)_{n=0}^{\infty}& eeimg=&1&&，即&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bn+%5Crightarrow+%5Cinfty%7D%7Ba_n%7D+%3D0& alt=&\textstyle \lim_{n \rightarrow \infty}{a_n} =0& eeimg=&1&&，序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=0& alt=&0& eeimg=&1&&，且对于每个&img src=&///equation?tex=n& alt=&n& eeimg=&1&&，&img src=&///equation?tex=f%28a_n%29%5Cneq+0& alt=&f(a_n)\neq 0& eeimg=&1&&，因为&img src=&///equation?tex=a_n%5Cin+E%5Cbackslash%5C%7Bx_0%5C%7D& alt=&a_n\in E\backslash\{x_0\}& eeimg=&1&&。又因为&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+%5Cmathbf%7BR%7D%5Csetminus%5C%7B0%5C%7D%7D%7B%5Cfrac%7B%5Csin+x%7D%7Bx%7D+%7D+%3D1& alt=&\textstyle \lim_{x \rightarrow 0;x\in \mathbf{R}\setminus\{0\}}{\frac{\sin x}{x} } =1& eeimg=&1&&，而序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=0& alt=&0& eeimg=&1&&且每一项都不为&img src=&///equation?tex=0& alt=&0& eeimg=&1&&，即对每个&img src=&///equation?tex=n& alt=&n& eeimg=&1&&，&img src=&///equation?tex=f%28a_n%29%5Cin+%5Cmathbf%7BR%7D%5Cbackslash%5C%7B0%5C%7D& alt=&f(a_n)\in \mathbf{R}\backslash\{0\}& eeimg=&1&&，于是序列&img src=&///equation?tex=%28g%28f%28a_n%29+%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(g(f(a_n) )_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=1& alt=&1& eeimg=&1&&，也就是序列&img src=&///equation?tex=%5Ctextstyle+%5Cleft%28%5Cfrac%7B%5Csin%28a_n%5Csin%5Cfrac%7B1%7D%7Ba_n%7D%29+%7D%7Ba_n%5Csin%5Cfrac%7B1%7D%7Ba_n%7D%7D+%5Cright%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&\textstyle \left(\frac{\sin(a_n\sin\frac{1}{a_n}) }{a_n\sin\frac{1}{a_n}} \right)_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=1%0A& alt=&1
& eeimg=&1&&，因此函数&img src=&///equation?tex=%5Ctextstyle+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+& alt=&\textstyle \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} & eeimg=&1&&在&img src=&///equation?tex=0& alt=&0& eeimg=&1&&处沿着&img src=&///equation?tex=E& alt=&E& eeimg=&1&&收敛到&img src=&///equation?tex=1& alt=&1& eeimg=&1&&，即&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+E%5Cbackslash%5C%7B0%5C%7D%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+%3D1& alt=&\textstyle \lim_{x \rightarrow 0;x\in E\backslash\{0\}} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} =1& eeimg=&1&&&br&&br&注意，下面命题&b&不成立&/b&&br&&b&命题. &/b&（&b&不成立）&/b&设&img src=&///equation?tex=X%2CY& alt=&X,Y& eeimg=&1&&是&img src=&///equation?tex=%5Cmathbf%7BR%7D& alt=&\mathbf{R}& eeimg=&1&&的子集，设&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&是&img src=&///equation?tex=X& alt=&X& eeimg=&1&&的极限点，&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&是&img src=&///equation?tex=Y& alt=&Y& eeimg=&1&&的极限点，设&img src=&///equation?tex=f%3AX%5Cto+Y& alt=&f:X\to Y& eeimg=&1&&是函数，使得&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+x_0%3Bx%5Cin+X%5Cbackslash%5C%7Bx_0%5C%7D%7D%7Bf%28x%29%7D+%3Dy_0& alt=&\textstyle \lim_{x \rightarrow x_0;x\in X\backslash\{x_0\}}{f(x)} =y_0& eeimg=&1&&&br&，设&img src=&///equation?tex=g%3AY%5Cto+%5Cmathbf%7BR%7D& alt=&g:Y\to \mathbf{R}& eeimg=&1&&是函数，使得&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7By+%5Crightarrow+y_0%3Bx%5Cin+Y%5Cbackslash%5C%7By_0%5C%7D%7D%7Bg%28y%29%7D+%3Dz_0& alt=&\textstyle \lim_{y \rightarrow y_0;x\in Y\backslash\{y_0\}}{g(y)} =z_0& eeimg=&1&&，那么&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+x_0%3Bx%5Cin+X%5Cbackslash%5C%7Bx_0%5C%7D%7D%7Bg%28f%28x%29%29%7D+%3Dz_0& alt=&\textstyle \lim_{x \rightarrow x_0;x\in X\backslash\{x_0\}}{g(f(x))} =z_0& eeimg=&1&&&br&这个命题是&b&错误&/b&的，因此不能直接根据&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D+%7D+%3D0& alt=&\textstyle \lim_{x \rightarrow 0}{x\sin\frac{1}{x} } =0& eeimg=&1&&和&img src=&///equation?tex=+%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%7D%7B%5Cfrac%7B%5Csin+x%7D%7Bx%7D+%7D+%3D1& alt=& \textstyle \lim_{x \rightarrow 0}{\frac{\sin x}{x} } =1& eeimg=&1&&来说明&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+0%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+%3D1& alt=&\textstyle \lim_{x \rightarrow 0} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} =1& eeimg=&1&&&br&&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+x_0%3Bx%5Cin+X%5Cbackslash%5C%7Bx_0%5C%7D%7D%7Bg%28f%28x%29%29%7D& alt=&\textstyle \lim_{x \rightarrow x_0;x\in X\backslash\{x_0\}}{g(f(x))}& eeimg=&1&&不一定等于&img src=&///equation?tex=z_0& alt=&z_0& eeimg=&1&&。但如果函数&img src=&///equation?tex=g& alt=&g& eeimg=&1&&在&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&处连续，这个命题是成立的。因此可以通过补充定义&img src=&///equation?tex=%5Ctfrac%7B%5Csin%28x%29%7D%7Bx%7D& alt=&\tfrac{\sin(x)}{x}& eeimg=&1&&在&img src=&///equation?tex=0& alt=&0& eeimg=&1&&处的值使得它连续，这样就可以使用上述命题说明&img src=&///equation?tex=%5C%28%5Clim_%7Bx+%5Crightarrow+0%3Bx%5Cin+E%5Csetminus%5C%7B0%5C%7D%7D+%5Cfrac%7B%5Csin%28x%5Csin%5Cfrac%7B1%7D%7Bx%7D%29%29+%7D%7Bx%5Csin%5Cfrac%7B1%7D%7Bx%7D%7D+%3D1%5C%29& alt=&\(\lim_{x \rightarrow 0;x\in E\setminus\{0\}} \frac{\sin(x\sin\frac{1}{x})) }{x\sin\frac{1}{x}} =1\)& eeimg=&1&&。&br&&br&因为对于每个完全由&img src=&///equation?tex=X%5Cbackslash%5C%7B0%5C%7D& alt=&X\backslash\{0\}& eeimg=&1&&的元素组成，并且收敛到&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&的序列&img src=&///equation?tex=%28a_n%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(a_n)_{n=0}^{\infty}& eeimg=&1&&，序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&，但序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&可能有&b&无限&/b&多项等于&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&，而&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7By+%5Crightarrow+y_0%3By%5Cin+Y%5Cbackslash%5C%7By_0%5C%7D%7D%7Bg%28y%29%7D+%3Dz_0& alt=&\textstyle \lim_{y \rightarrow y_0;y\in Y\backslash\{y_0\}}{g(y)} =z_0& eeimg=&1&&只能说明对于每个完全由&img src=&///equation?tex=Y%5Cbackslash%5C%7B0%5C%7D& alt=&Y\backslash\{0\}& eeimg=&1&&的元素组成，并且收敛到&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&的序列&img src=&///equation?tex=%28b_n%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(b_n)_{n=0}^{\infty}& eeimg=&1&&，序列&img src=&///equation?tex=%28g%28b_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(g(b_n))_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=z_0& alt=&z_0& eeimg=&1&&，当序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&有无限多项等于&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&，我们不能推出序列&img src=&///equation?tex=%28g%28f%28a_n%29%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(g(f(a_n)))_{n=0}^{\infty}& eeimg=&1&&收敛到&img src=&///equation?tex=z_0& alt=&z_0& eeimg=&1&&，但下面两种情况可以&br&&ul&&li&对于每个收敛到&img src=&///equation?tex=x_0& alt=&x_0& eeimg=&1&&的序列&img src=&///equation?tex=%28a_n%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(a_n)_{n=0}^{\infty}& eeimg=&1&&，序列&img src=&///equation?tex=%28f%28a_n%29%29_%7Bn%3D0%7D%5E%7B%5Cinfty%7D& alt=&(f(a_n))_{n=0}^{\infty}& eeimg=&1&&只有有限多项等于&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&，那么就有&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+x_0%3Bx%5Cin+X%5Cbackslash%5C%7Bx_0%5C%7D%7D%7Bg%28f%28x%29%29%7D+%3Dz_0& alt=&\textstyle \lim_{x \rightarrow x_0;x\in X\backslash\{x_0\}}{g(f(x))} =z_0& eeimg=&1&&&br&&/li&&li&函数&img src=&///equation?tex=g%28y%29& alt=&g(y)& eeimg=&1&&在&img src=&///equation?tex=y_0& alt=&y_0& eeimg=&1&&处&b&连续&/b&时，我们也有&img src=&///equation?tex=%5Ctextstyle+%5Clim_%7Bx+%5Crightarrow+x_0%3Bx%5Cin+X%5Cbackslash%5C%7Bx_0%5C%7D%7D%7Bg%28f%28x%29%29%7D+%3Dz_0& alt=&\textstyle \lim_{x \rightarrow x_0;x\in X\backslash\{x_0\}}{g(f(x))} =z_0& eeimg=&1&&。&/li&&/ul&
谢邀。以前回答过这个问题：，不过那个问题好像被举报了。下面是以前的回答，修订了一下错误。 是的，不过更准确的应该是\(\lim_{x \rightarrow 0;x\in E\setminus\{0\}} \frac{\sin(x\sin\frac{1…
我觉得你老师的意思是，x_k=1/(kpi)的时候分母是0，式子没定义，而x_k是一个收敛到0的序列，所以极限不是well-defined的。&br&个人觉得这个解释make sense，但是问题本身属于没什么价值的细枝末节，没太多纠结的必要。
我觉得你老师的意思是，x_k=1/(kpi)的时候分母是0，式子没定义，而x_k是一个收敛到0的序列，所以极限不是well-defined的。 个人觉得这个解释make sense，但是问题本身属于没什么价值的细枝末节，没太多纠结的必要。
已有帐号？
无法登录？
社交帐号登录
白痴一只。}
我要回帖
更多推荐
- ·什么是网易云音乐黑胶黑胶vip？
- ·如何安装广联达软件都有什么软件啊？
- ·解放j6p解放j6l原车柴暖如何开启的一下不着了,后来关了从开启动不了电源灯一直闪烁是什么情况啦？
- ·2021年体育生可以考哪些大学及分数线专业的录取分数是多少呢？
- ·网易云svip和vip的区别黑胶VIP和畅听VIP有什么区别啊？
- ·树林（两个字的偏旁相同中有一个字的两部分由相同偏旁组成）：
- ·从北京到美国北达科他州大学，怎样做飞机最便宜
- ·out of sight ,out of mind 是mind什么意思思啊？
- ·20℃时，食盐的溶解度是36g，将40g食盐放入100g水中，所得食盐什么是饱和溶液液的质量是（ ）g。
- ·核力与洛伦兹力习题之间的关系
- ·按图索骥，日本地名 英文
- ·素质报告册家长的话里面的写心里话200字（关于学习） 急急急 好的给300分啊！！
- ·广州卫星天线安装装方法？
- ·" If one day i walked upand you don't be certain to wait me . "是什么意思？
- ·小学四年级寒假日记作业希望别人原谅自己,你可以说什么
- ·关于 大东文化大学 东松山湖钓鱼文化网キャンパス
- ·彩灯上串联着20只小灯泡，如果电流最接近200ma为200mA，则电流最接近200ma为？
- ·问个用定积分求极限限的题
- ·求一篇格式规范的 商务管理 本科毕业论文格式。。谢谢！！！
- ·847亿kW·h可以文员是做什么的
- ·帮忙翻译简历下，
- ·英语题，英语短文改错方法
- ·中山大学选课系统 体育选课 四个都选还会落选吗
- ·英语常用短语中常用__and__表示甜言蜜语
- ·本人求自贡市卫生局中考英语播放的钢琴曲，哪位大虾知道此曲名字？
- ·许晴东边日出西边雨雨，道是无情却有晴。是什么修辞手法
- ·甲类车间的配电室和消防控制室门是否允许设在车间内？
- ·2011河北政法干警考试为什么一直不公布名单和开学日期？知道的进，不知道的绕行，别胡说！
- ·现在的痕量烃分析仪灌溉管价格是多少？
- ·怎样能取得好成绩个好成绩啊？
- ·一年级语文初二上册数学测试题 第二单元测试题 急求
- ·求2篇800字高中材料作文及范文，急，请看材料，谢谢！！！
