Can SiNx + cosx be replaced by X + 1 equivalently when x approaches 0? Can't addition and subtraction replace infinitesimal equivalently? Be sure. Can we replace the limit? How to find the limit of 1 / x power of this formula?

Can SiNx + cosx be replaced by X + 1 equivalently when x approaches 0? Can't addition and subtraction replace infinitesimal equivalently? Be sure. Can we replace the limit? How to find the limit of 1 / x power of this formula?


Addition and subtraction can not be replaced equivalently. It is said that if the whole of addition and subtraction is replaced, sometimes it is OK. The key depends on whether it is equivalent infinitesimal, that is, whether the replaced factor and the replaced factor are equivalent infinitesimal
For example, in this question, can SiNx + cosx be replaced by 1 + X? The judgment method is to divide the two and find the limit. If the limit value is 1, it depends on the situation. Sometimes it is OK to replace it as a whole
SiNx + cosx and 1 + X are equivalent factors, but whether they can be replaced depends on the situation
For example, your problem is the infinity limit of 1. Generally, we don't and can't use the equivalent infinitesimal in the base of index. There are two ways to solve this problem: one is to take logarithms on both sides, and then find the limit; the other is to change it into the exponential form with E as the base, and then find the limit
Original formula = (SiNx + cosx) ^ 1 / x = (1 + SiNx + cosx-1) ^ 1 / (SiNx + cosx-1) * (SiNx + cosx-1) * 1 / X
For (1 + SiNx + cosx-1) ^ 1 / (SiNx + cosx-1), we use two important limits and get that the limit is e
There are many ways to find the limit of (SiNx + cosx-1) * 1 / x, such as Taylor series expansion (expansion to the first order of X is OK, and the rest is replaced by O (x)), (I forgot the expansion) the limit is 1
So the answer is e



Given that the point (3, 5) is on the straight line y = ax + B (a, B are constants, and a ≠ 0), then the value of ab − 5 is______ .


∵ point (3, 5) is on the straight line y = ax + B, ∵ 5 = 3A + B, ∵ B-5 = - 3a, then ab − 5 = a − 3A = − 13



Given that the point (2,6) is on the straight line y = ax + B (a, B are constants, and a is not equal to 0), then the value of a in B-6 is


Substituting point (3,5) into y = ax + B, we get: 5 = 3A + B, that is: B-5 = - 3a, so: a of B-5 = - 1 / 3