Original Address: https://zhaokaifeng.com/?p=1917
topic
When (x \ rightarrow 0 \) \ time, \ (F (X) = X-\ SiN AX \) and \ (g (x) = x ^ {2} \ ln (1-bx) \) is equivalent to infinitesimal, then ()
( A ) \(a=1,b=-\frac{1}{6}.\)
( B ) \(a=1,b=\frac{1}{6}.\)
( C ) \(a=-1,b=-\frac{1}{6}.\)
( D ) \(a=-1,b=\frac{1}{6}.\)
Resolve
Because of \ (f (x) \) and \ (g (x) \) is equivalent infinitesimal, and therefore, according to the "infinitesimal Comparative" Equivalent infinitesimal theorems about:
设 \(\lim \alpha(x)=0, \lim \beta(x)=0,\)
If \ (\ Lim \ FRAC {\ Alpha (X)} {\ Beta (X)} =. 1 \) , then \ (\ alpha (x) \ ) and \ (\ beta (x) \ ) is equivalent infinitesimal , referred to as \ (\ Alpha (X) \ SIM \ Beta (X) \) .
Therefore, we have:
\(\lim_{x \rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{x-\sin ax}{x^{2}\ln(1-bx)}=1.\)
In the "common equivalent infinitesimal" while and (\ sin x \) \ and (X \) \ infinitesimal about two equivalents, as follows:
\(\sin x \sim x;\)
\(x-\sin x \sim \frac{1}{6}x^{3}.\)
At the same time, and \ (\ ln x \) and \ (x \) related to the equivalent infinitesimal also has two, as follows:
\ (\ Ln (1 + x) \ sim x; \)
\(x-\ln(1+x)\sim \frac{1}{2}x^{2}.\)
So, the question we now need to consider is: which of the two need to be combined equivalent infinitesimal simplification original style?
Here select and determine which of the two infinitesimal equivalent basis is given the title of "equivalent infinitesimal" use. That is, during the original equation simplified operation, it is necessary to ensure that the numerator and denominator of each other equivalent infinitesimal, every step must follow this principle, and finally result in simplification out of the numerator and denominator must be mutually equivalent infinitesimal only in this way can be equated and original style.
We know from the previous calculation, the original type of molecule is:
\(x-\sin ax\)
Original denominator is:
\(x^{2}\ln(1-bx)\)
So, effectively simplified form of the molecule has the following four:
\(x-\sin ax=x-ax\) (1)
OR
\ (x- \ sin ax = \ sin x- \ sin ax \) (2)
OR
\(x-\sin ax=x-[ax-\frac{1}{6}(ax)^{3}]=x-ax+\frac{1}{6}a^{3}x^{3}\) (3)
OR
\(x-\sin ax=\frac{1}{6}x^{3}+\sin x-\sin ax\) (4)
Effective simplification form the denominator of the following two:
\(x^{2}\ln(1-bx)=x^{2}(-bx)=-bx^{3}\) (5)
OR
\(x^{2}\ln(1-bx)=x^{2}[(-bx)-\frac{1}{2}(-bx)^{2}]=-bx^{3}-\frac{1}{2}b^{2}x^{4}\) (6)
Due to ensure that every step of the calculation process, the numerator and denominator are equivalent infinitesimal, therefore, we must first look at those equations can be combined to form an equivalent infinitesimal.
(1) to (6) six formulas variable \ (X \) is the number of power as follows:
(1): 1 contains only power;
(2): 1 contains only power;
(3): 3 comprising a power and power;
(4): comprising a power and a power of 3;
(5): contains only 3 power;
(6): comprising 3 and 4 th power.
Since the denominator corresponding to (5) and (6) contain two formulas third power, the corresponding molecules (1) and (2) a change in any case does not appear to the third power, equivalent to not constituting denominator infinitesimal, and therefore excluded. Further, (4) have the formula \ (\ sin x \) and \ (\ SiN AX \) , and the denominator is no corresponding form, so equation (4) is substantially excluded.
Now the rest of the corresponding molecule (3) and a denominator corresponding to the formula (5) and (6) a. Since (6) contained in the formula \ (X \) is the fourth power, and (3) does not occur in any case change fourth power, and therefore, the reduction process is to be correct in (3) and (5) where production.
Based on the above analysis, try to simplify as follows:
\(原式=\lim_{x\rightarrow 0}\frac{x-ax+\frac{1}{6}a^{3}x^{3}}{-bx^{3}}=\lim_{x\rightarrow0}\frac{(1-a)x+\frac{1}{6}a^{3}x^{3}}{-bx^{3}}\)
The denominator is not a power, therefore, in order to ensure "the numerator and denominator of each other equivalent infinitesimal" This condition is always set up, the only way is to make \ (1-A = 0 \) , followed, according to \ (f (x) \ sim g (x) \) numerator and denominator of the obtained correspondence relationship, we can get:
\(\frac{1}{6}a^{3}=-b\)
Two simultaneous type:
\(\left\{\begin{matrix}1-a=0,\\ \frac{1}{6}a^{3}=-b.\end{matrix}\right.\)
Solutions have to:
\(\left\{\begin{matrix}a=1,\\ b=-\frac{1}{6}.\end{matrix}\right.\)
To sum up, this question is the correct option: A
By this question, we can conclude the following rules using the original formula equivalent infinitesimal process of simplification:
Note the original type the numerator and denominator of the small type infinite (equivalent to higher-order, low-level, the same order, K-order), the calculation process, we need to infinitesimal type is always consistent for all premise computing;
Use common equivalent infinitesimal Simplification time are generally from complex to simple, that is simplification of trends are possible only appear in the formula \ (the X-\) , for example, \ (\ sin x \) into \ (X \) , the \ (\ ln (1 + x ) \) into \ (X \) and the like.
Further, the formulas and into another part of the same type are more likely to form part of a simplified calculation, for example, in the present problem, the denominator is the \ (X ^ {2} \ LN (. 1-BX) \) , put \ (\ ln (1-bx) \) into \ (- bx \) obviously makes the equation in the form of a more unified, more conducive to calculate the back;
Simplification process should strictly follow the formula, with particular attention to the negative sign in front of the variables and parameters, first parentheses maintain its original form, if necessary, after a calculated step by step.