Find the limit, when x tends to 0, find the limit of sinx x arc sinx x tanx arc tanx x

x 0, then sinx x arcsinx tanx [between them is equivalent infinitesimal at x 0.]

lim(x→0）（sinx/x+arc sinx/x+tanx/x+arc tanx/x）

lim(x→0)(sinx/x)+lim(x→0)(arcsinx/x)+lim(x→0)(tanx/x)+lim(x→0)(arctanx/x)

lim(x→0)(x/x)+lim(x→0)(x/x)+lim(x→0)(x/x)+lim(x→0)(x/x)

Appendix Common Equivalent Infinitesimals].

x→0，x~sinx~tanx~arcsinx~arctanx~(e^x)-1~ln(x+1)

1-cosx)~x²/2

[sinx+arc sinx+tanx+arc tanx] x brings x=0 to get 0 0

Lobida, up and down the same guidance.

cosx+1 root number (1-x 2)+sec 2 x+1 (1+x 2)] 1

Bring x=0 in.

Get [1+1+1+1]=4

The limit is 4

Because x tends to 0, x sinx tanx arcsinx arctanx ( is an equivalence sign) so sinx x=arc sinx x=tanx x=arc tanx x=1, so the original equation is equal to 4

x 0, then sinx arcsinx tanx [between them is equivalent infinitesimal under x 0] lim(x 0) ( x 0) ( x 0 ) ( x 0 x + arc sinx x + tan x + arc tanx x) = lim (x 0) (sinx x) + lim (x 0) (arc sinx x) + lim (x 0) (tanx x) + lim (x 0) (arc tanx x) = lim (x 0) (x x...).

x 0lim (x-tanx) (x 2*sinx) The limit is 0 0 type, using l'hospital rule = lim (x-tanx).' / (x^2*sinx)'=lim (1-1/cos^2x) / (2xsinx+x^2cosx)=lim (-sin^2x/cos^2x) / (2xsinx+x^2cosx)=lim (-sin^2x/cos^2x)/x^2 / (2xsinx+x...

The molecule tanx-sinx tanx(1-cosx), tanx is equivalent to x, 1-cosx is equivalent to 1 2*x*x, and the denominator sinx is equivalent to x, so the original limit lim tanx(1-cosx) sinx's three-plexed pure sub-shaded Zheng Difang macro nucleus lim (x*1 2*x*x) x*x*x) 1 2

tan3x 5x limit 3 5 when x is trending 0

tan mx sin nx when the huidong jan x twitch front is close to the limit m nx 0. tanx～sinx ln(1+x)～arctanx～arcsinx

Equivalent front pants.