x→arcs inx； x→tanx； x→arctanx； x→ln( 1+x)； x→(e^x- 1)；

[( 1+x)^n- 1]→nx； ( 1-cosx)→x * x/2； a^x- 1→xlna，ln( 1+x)→x； So is mclaughlin's formula,

That symbol is difficult to write. You can find it in textbooks or exercises. For example: 1 limx→0tanx-sinxx3.

Give a few examples with infinitesimal.

Example 1 limx→0tanx-sinxx3

Solution: The original formula = limx → 0 sinx (1-cosx) x3 cosx = limx → 0x12x2x3 (∵ sinx ~ x, 1-cosx ~ x22) = 12.

This problem can also be solved by Robita's law, but not by property ④.

∵ tanx-sinxx3=x-xx3=0, which does not meet the condition of property ④, otherwise the wrong conclusion is 0.

Example 2 limx→0e2x-3 1+xx+sinx2

Solution: The original formula = limx → 0e2x-1-(31+x-1) x+x2 = limx → 02x-13xx (1+x) = 53.

Example 3 limx→0( 1x2-cot2x)

Solution 1: original formula = LIMX→0 sin 2 x-x2 cos 2 x2 sin 2 x.

= limx→0(sinx+xcosx)(sinx-xcosx)x4

= limx→0x 2( 1+cosx)( 1-cosx)x4(∵sinx ~ x)

= limx→0( 1+cosx)( 1-cosx)x2

= limx→0 12 x2( 1+cosx)x2 = 1

Solution 2: Original formula = limx→0 tan2x-x2 tan2x

=limx→0(tanx+x)(tanx-x)x4

=limx→02x(tanx-x)x44 (∵ tanx~x)

=limx→02(tanx-x)x3

=limx→02(sec2x- 1)3x2

=23limx→0tan2xx2=23 (∵ tanx~x)

Example 4 [3] limx → 0+tan (sinx) sin (tanx)

Solution: Original formula = limx → 0+sec2 (sinx) cosx2tan (sinx) cos (tanx) sec2x2sin (tanx) (according to Robita's law).

= limx → 0+sec2 (sinx) cosxcos (tanx) sec2xlimx → 0+sin (tanx) tan (sinx) (separation of non-zero limit product factors)

=limx→0+sin(tanx)tan(sinx) (calculate non-zero limit)

= limx → 0+cos (sinx) sec2x2sin (tanx) sec2 (sinx) cosx2tan (sinx) (according to Robita's law)

= limx→0+cos(sinx)sec 2 xsec 2(sinx)cosx limx→0+tan(sinx)sin(tanx)

=limx→0+tan(sinx)sin(tanx)

There is a cycle, and at this time, the result cannot be obtained by Robita's law. What shall we do? Replace it with equivalent infinitesimal.

∫x ~ sinx ~ tanx(x→0)

∴ Original formula =limx→0+xx= 1 and get the solution.