Higher numbers, on equivalent infinitesimals LIM (x approaches 0) 1 / (1-cosx) + 1 / TaNx Can TaNx, (1-cosx) be replaced by the equivalent infinitesimal? If not, why not, can not all multiplication and division be used? 2. LIM (x approaches infinity) [E / (1 + 1 / x) ^ x] ^ x Can (1 + 1 / x) ^ X be replaced by e? If not, why not?

Higher numbers, on equivalent infinitesimals LIM (x approaches 0) 1 / (1-cosx) + 1 / TaNx Can TaNx, (1-cosx) be replaced by the equivalent infinitesimal? If not, why not, can not all multiplication and division be used? 2. LIM (x approaches infinity) [E / (1 + 1 / x) ^ x] ^ x Can (1 + 1 / x) ^ X be replaced by e? If not, why not?

The two problems are actually the same problem
It is a quantity in the form of a factor. Note that it is relative to the whole expression in the form of a factor,
Not a single part is in the form of a factor
For example, in the first question, 1-cosx is a factor in the first part, but it is not a factor relative to the whole expression,
So it can't be replaced equivalently
Lim 1 / (1-cosx) + Lim 1 / TaNx, 1-cosx is a factor and can be replaced
Of course, it's wrong to do so, because it can't be written in the above form
The correct way is to divide first, and then use lobita's rule or Taylor's expansion
Similarly, (1 + 1 / x) ^ x is not a factor relative to the whole expression, so it cannot be replaced equivalently
The correct way is to take logarithm first, and then use lobita's law or Taylor's expansion
Taylor expansion. After taking logarithm,
lim x*(1－xln(1+1/x))
= Lim x * (1-x [1 / X-1 / (2x ^ 2) + small o (1 / x ^ 2)]
= Lim x * (1 / 2x + small o (1 / x))
＝1/2,
So the original limit is e ^ (1 / 2)