Because (1+x 2)- 1 under the radical sign is equivalent to 1/2 * x 2,

So lim (1/2 * x 2)/x = lim1/2 * x = 0.

2. The order of infinitesimal is: (1), (2), (3), (4) and (6).

Arcsinx is equivalent to X, so arctanx is equivalent to X, and In (1+X) is equivalent to X. 。

Therefore, (1), (3) and (4) are all infinitesimals of the same order and are equivalent infinitesimals.

And lim[ under the root sign (1+tanx)-under the root sign (1-sinx) ]]/x x.

(It can be solved by multiplying the numerator and denominator by the root sign (1+tanx)+ root sign (1-sinx) at the same time. )

= lim [1+tanx-(1-sinx)]/[x * (1+tanx)+under the root sign (1-sinx))]

= lim (tanx+sinx)/[x * (1+tanx)+under the radical sign (1-sinx))]

= lim (tanx+sinx)/x * lim1/((kloc-0/+tanx)+root sign (1-sinx))

=lim(tanx+sinx)/x * 1/2

= 1/2lim(tanx/x+sinx/x)

= 1/2 *( 1+ 1)

= 1

Therefore, (2) is the same order infinitesimal of X and the equivalent infinitesimal.

because

lim(cscx-cotx)/x

=lim( 1-cosx)/(x*sinx)

=lim( 1-cosx)/(x^2)

=limsinx/2x

= 1/2

Therefore, (6) is the same order infinitesimal.

3. The order of infinitesimal is (1), (2), (3) and (4).