Original: https://zhaokaifeng.com/?p=1935
topic
The following proposition is correct ()
( A ) 若 \(\lim_{x \rightarrow x_{0}} f(x) \geqslant \lim_{x \rightarrow x_{0}}g(x)\), 则 \(\exists \varepsilon > 0\), 当 \(0<|x-x_{0}|<\varepsilon\) 时，\(f(x) \geqslant g(x)\).
(B) if \ (\ EXISTS \ varepsilon> 0 \) , when the \ (0 <| x-x_ {0} | <\ varepsilon \) when, \ (F (X)> G (X) \) , and \ (\ lim_ {X \ rightarrow X_ {0}} F (X) = A_ {0}, \ lim_ {X \ rightarrow X_ {0}} G (X) = B_ {0} \) , then \ (A_ } 0 {> 0} {B_ \) .
( C ) 若 \(\exists \varepsilon>0\), 当 \(0<|x-x_{0}|<\varepsilon\) 时，\(f(x)>g(x)\), 则 \(\lim_{x \rightarrow x_{0}}f(x) \geqslant \lim_{x \rightarrow x_{0}}g(x)\).
( D ) 若 \(\lim_{x \rightarrow x_{0}}f(x)>\lim_{x \rightarrow x_{0}}g(x)\), 则 \(\exists \varepsilon>0\), 当 \(0<|x-x_{0}|<\varepsilon\) 时，\(f(x)>g(x)\).
Resolve
The concept study title is a postgraduate mathematics class more difficult questions, such practice is that in addition to the difficulty to pull tight concept, how much space is free to play solver no. Moreover, the nuances of the concept study questions are examined concept, an inattentive may examine the wrong question.
As can be seen from the four options to this question, the focus of this title examines the limits of this part of the function. More detailed, the present title examines the definition of the limit function when \ (x \ rightarrow x_ {0 } \) defined time limit, as follows:
Known \ (\ lim_ {x \ rightarrow x_ {0}} f (x) = A \)
Given any \ (\ varepsilon> 0 \) , there is a positive number \ (\ Delta \) , when the \ (0 <x-x_ { 0} <\ delta \) when there \ (| f (x) -A | <\ varepsilon \) .
NOTE: The above definition Popular said point is, when the \ (X \) and \ (x_ {0} \) close enough, \ (f (x) \) and \ (f (x) \) limit \ (A \) is also close enough.
This problem also examines the "security number" of the nature of the limit function, as follows:
Set \ (\ Lim F (X) = A> 0 \) , are within the limiting jurisdiction, \ (F (X)> 0 (F (X)> \ FRAC {A} {2}) \) .
Conversely, \ (F (X)> 0 \) and \ (\ Lim F (X) = A \ Rightarrow A \ geqslant 0 \) .
NOTE: When \ (\ x rightarrow x_ {0 } \) , the "range limit jurisdiction" refers to \ (x_ {0} \) of the coring neighborhood; when \ (x \ rightarrow \ infty \ ) when "limit the scope of jurisdiction" refers to infinity.
For security number of the limits of nature function, we need to clear the following points:
No. Paul answers questions the premise of "extreme functions involved are present", which is the premise to solve all the issues involved limits: To study and use limit, the limit must be present;
Security number of all local security number, i.e. only the presence of the security number in the range of limits jurisdiction;
Is greater than the limit \ (0 \) can be introduced function is greater than \ (0 \) , not Release function is equal to \ (0 \) or function less than \ (0 \) . Greater than by the function \ (0 \) can be introduced limit greater than \ ( 0 \) or equal to the limit \ (0 \) , but only what is greater than the uncertainty limit \ (0 \) or only less than (0 \) \ case, the write limit greater than or equal to \ (0 \) in the form of .
The following is an analysis of this question in each option.
A Options
This option gives:
\(\lim_{x \rightarrow x_{0}} f(x) \geqslant \lim_{x \rightarrow x_{0}}g(x)\)
This shows that \ (f (x) \) and \ (g (x) \) limit exists (the limit to meet the premise of research issues, conditions available, can continue to ponder the next step) and \ (f (x ) \) is not less than the limit \ (f (x) \) limit.
Thus, we have:
\(\lim_{x \rightarrow x_{0}}(f(x)-g(x)) \geqslant 0\)
The next option is given:
If \ (\ EXISTS \ varepsilon> 0 \) , when the \ (0 <| x-x_ {0} | <\ varepsilon \) when
This shows that we are to "within the jurisdiction of the limit of function," saying discussion of this option, with the premise of using security number of conditions are available, can continue to ponder the next step.
The next option is noted from the above conditions can be introduced \ (f (the X-) \ geqslant G (the X-) \) .
This conclusion is wrong. For the following reasons:
If the function \ (f (x) \) limit \ (A> 0 \) , then the function can be introduced \ (F (X)> 0 \) ;
If the function \ (f (x) \) limit \ (A <0 \) , then the function can be introduced \ (F (X) <0 \) ;
If the function \ (f (x) \) limit \ (A = 0 \) , it can not determine the function \ (f (x) \) is greater than \ (0 \) , less than \ (0 \) is equal to \ ( 0 \) . the reason is that if \ (a = 0 \) we do not know the function \ (f (x) \) is greater than \ (0 \) tends to limit the direction of \ (a \) , or in less than \ (0 \) approach in the direction of the limit \ (a \) , or whether \ (F (X) = 0 \) .
As shown in FIG. 1, when the limit is equal to the function \ (0 \) , the functions may be greater than \ (0 \) is:
1. FIG partial image y = 1 / x, and generated using www.desmos.com
2, when the function limit equal to 0, the function may be smaller than \ (0 \) is:
FIG 2. y = 1 / (- x) of partial images generated using www.desmos.com
The third case, when the function limit equal to \ (0 \) , the function may also be equal to \ (0 \) , as shown below:
图 3. y=0 的局部图像，使用 www.desmos.com 生成
因此，已知极限 \(\lim_{x \rightarrow x_{0}}[f(x)-g(x)]\geqslant0\), 并不能推导出函数 \(F(x)=[f(x)-g(x)]\geqslant0\).
综上可知，选项 A 是错误的。
B 选项
题目中给出了如下条件：
若 \(\exists \varepsilon>0\), 当 \(0<|x-x_{0}|<\varepsilon\) 时
因此，本题符合函数极限保号性的使用条件，条件可用，可以继续接下来的思考步骤。
接着，该选项给出：
\(f(x)>g(x)\)
于是，当我们令 \(F(x)=f(x)-g(x)\) 时，可以得出如下结论：
\(F(x)>0\)
接着，该选项又给出：
\(\lim_{x \rightarrow x_{0}}f(x)=A_{0}, \lim_{x \rightarrow x_{0}}g(x)=B_{0}\)
这说明函数 \(f(x)\) 和函数 \(g(x)\) 都是存在极限的，符合我们研究函数极限问题的大前提，条件可用，可以继续接下来的思考步骤。
最后，该选项给出了他的结论：
\(A_{0}>B_{0}\)
有了这个结论，结合前面的条件，我们可以把该选项改写成如下形式：
已知函数 \(F(x)\) 存在极限，且函数 \(F(x)>0\), 则 \(\lim_{x \rightarrow x_{0}}F(x)>0\).
这个结论显然是错误的，因为已知函数大于 \(0\) 的时候，其极限是可能等于 \(0\) 的，例如对 A 选项的解析中给出的图 1, 函数 \(f(x)=\frac{1}{x}\) 始终是大于 \(0\) 的，但是其极限却是等于 \(0\) 的。
综上可知，选项 B 是错误的。
C 选项
该选项的错误比较明显，因为选项中没有指明函数 \(f(x)\) 和函数 \(g(x)\) 的极限存在，缺少了研究极限问题的大前提，那么，接下来的所有说明和结论都是没有根据也没有意义的。不过，如果 C 选项像 B 选项一样指明函数 \(f(x)\) 和函数 \(g(x)\) 的极限是存在的，那么该选项的表述就是正确的，原因在 B 选项中已经分析过。
综上可知，选项 C 是错误的。
D 选项
该选项首先给出了如下条件：
\(\lim_{x \rightarrow x_{0}}f(x)>\lim_{x \rightarrow x_{0}}g(x)\)
若我们令 \(F(x)=f(x)-g(x)\), 则上面的条件可以改写成：
\(\lim_{x \rightarrow x_{0}}F(x)>0\)
接着选项给出了：
若 \(\exists \varepsilon>0\), 当 \(0<|x-x_{0}|<\varepsilon\) 时
这说明我们是要在“函数极限的管辖范围内”讨论这个选项的说法，具备使用保号性的前提，条件可用，可以继续接下来的思考步骤。
接着，该选项给出了它的结论：
\(f(x)>g(x)\)
根据前面的分析可知，我们可以将此改写成：
\(F(x)>0\)
我们知道，当一个函数的极限存在且大于 \(0\) 的时候，在函数极限的管辖范围内，可以推导出该函数也大于 \(0\).
综上可知，选项 D 是正确的。
EOF