&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…
看了别人然后发现……还是不改了，，我就是永远不认真看题除非做不下去的那种根本不记得极限的定义了，，不过想象了一下可以用 x=1/(2n*pai+1), n趋向无穷大。 解决问题？我一开始没把x=1/k*pai当回事，因为这些都是实际例举出的数，讨论无穷小时可忽略。 又一想他也可趋向0似乎不太对啊，，我们不要x了还不行吗，，，，，==============还有觉得不要什么不会做就去matlab， mathematica（虽然我也会做个弊什么），基础题还是要会动笔算的吧sinx等价于x（x趋向于0）是没错，但x=0时，sinx=0=x你们是傻吗？！虽说“趋向于”不包括等于的情况，但还没讨论怎么就直接扣个“无意义”的帽子？为什么说sinx等价于x呢？（条件略）（ 记得同济六版说这个结论时用的是比值极限为1 ）与其用等价不如直接泰勒展开：sinx=x-x^3/6+....（读高数的都懂就这样了）只是当求sinx/x的极限时如同多项式分式求极限那样sinx展开的后面被忽略了（结果如此，过程略复杂）你高数老师要知道一定气死了（一定不会）。。算了我们看下题吧，第一眼就应该替换成求sint/t，（换元）我们当然要简化简化再简化。。一切逃不出基础。。（瞎扯的，但一看这分母分子一模一样不换元觉得好亏）sinx/x的图一定要知道（这好像是高中的吧喂）（大致如此，只要知道x向0、向无穷时的不同极限就可以了）安卓arity画的，这软件还不错。然后我们只要去求t是向0还是向无穷的就可以了（看上去也不是特殊值对吧）t=x*sin(1/x)看不出吧，换元：t=(1/u)*sin(u)=sin(u)/u，（u趋向无穷）得t趋向0得所求为1
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=&internal&&极限点&/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=&internal&&Dirichlet Function &/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。
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大二电气生，，并没有上过专业课