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Limit problem of higher mathematical function
1. In principle, it can be expanded after separation, and then infinitesimal is used for each score.

But this requires that the limit subtraction of the two formulas you separate is meaningful.

Not here.

Secondly, it is wrong to look at your equivalent infinitesimal.

tanx~x

sinx~x

Note that the denominator is (sinx) 3 ~ x 3.

because

The limit of tanx/(sinx) 3 ~ x/x 3 = 1/x 2 is positive infinity.

The limit of sinx/(sinx) 3 ~ x/x 3 = 1/x 2 is positive infinity.

Positive infinity-positive infinity is indefinite.

2. Taylor can also be directly expanded to a certain order (generally not used).

But because the order of denominator is x 3.

Your molecules must swell to at least x 3 to ensure no mistakes.

3. The correct approach:

tanx=sinx/cosx

Primitive up-down multiplication cosx

= (sinx-sinx cosx)/[(sinx) 3 Coase]

Divided by sinx (because the limit, x≠0, just tends to 0)

= (1-cosx)/[(sinx) 2 Coase]

At this time, use equivalent infinitesimal again.

1-cosx~x^2/2

sinx~x

cosx~ 1

=(x^2/2)/[x^2* 1]

= 1/2

So simplify it as much as possible first, and then the equivalent is infinitesimal. Note that only multiplication and division can use equivalent infinitesimal.