When x→0, the Taylor expansion of sinx is sinx = x+o (x).

O(x) means that the higher order of x is infinitesimal, so when x→0.

Can (sinx) ~ x when x→0 (sinx)? =x？ +o(x？ )

So when x→0, can (sinx)? ~x？ .

Example:

limx→0(sinx-tanx)/{[3√( 1+x^2)- 1][( 1+sinx)- 1]}

The denominator part can be replaced by infinitesimal "x 2/3" and "sinx/3".

If the denominator is substituted correctly, sinx/3 can be replaced by x/3. Molecules do this:

sinx-tanx=tanx(cosx- 1)~x*(-x^2/2)=-x^3/2(x->； 0)

So the final answer is lim {x->; 0}(-x^3/2)/(x^3/9)=-9/2.

x→0)sinx+(sinx)^2→0 1+sinx→( 1+sinx)^2( 1+sinx)^( 1/2)- 1→ 1+sinx- 1→sinx

When x is infinite, 1+sinx and 1+2 sinx+(sinx) 2 are very close.

The difference sinx+(sinx) 2 is infinitely small, so 1+2 sinx+(sinx) 2 is used instead of 1+sinx to get the square root (1+sinx)- 1.

Extended data

All equivalent infinitesimal formulas in advanced mathematics;

When x→0 and x≠0, then

x ~ sinx ~ tanx ~ arcsinx ~ arctanx

x~ln( 1+x)~(e^x- 1)；

( 1-cosx)~ x * x/2；

[( 1+x)^n- 1]~nx；

loga( 1+x)~ x/lna；

The x power of a ~ ~xlna; ;

(1/n power of (1+x) ~ 1/NX (n is a positive integer);

Note: it is a power, and ~ is equivalent, which I summed up when I did the problem.